10 ways to solve quadratic equations. Methods for solving quadratic equations

Antipyretics for children are prescribed by a pediatrician. But there are emergency situations with fever when the child needs to be given medicine immediately. Then the parents take responsibility and use antipyretic drugs. What is allowed to be given to infants? How can you lower the temperature in older children? What medications are the safest?

Kop'evskaya rural secondary comprehensive school

10 Ways to Solve Quadratic Equations

Head: Patrikeeva Galina Anatolyevna,

mathematic teacher

village Kopevo, 2007

1. History of the development of quadratic equations

1.1 Quadratic equations in Ancient Babylon

1.2 How Diophantus composed and solved quadratic equations

1.3 Quadratic equations in India

1.4 Quadratic equations by al-Khorezmi

1.5 Quadratic equations in Europe XIII - XVII centuries

1.6 About Vieta's theorem

2. Methods for solving quadratic equations

Conclusion

Literature

1. History of the development of quadratic equations

1.1 Quadratic equations in Ancient Babylon

The need to solve equations not only of the first, but also of the second degree, even in ancient times, was caused by the need to solve problems related to finding the areas of land plots and with excavation work of a military nature, as well as with the development of astronomy and mathematics itself. Quadratic equations could be solved around 2000 BC. e. Babylonians.

Using modern algebraic notation, we can say that in their cuneiform texts there are, in addition to incomplete ones, such, for example, complete quadratic equations:

X2 + X= ¾; X2 - X= 14,5

The rule for solving these equations, set out in the Babylonian texts, essentially coincides with the modern one, but it is not known how the Babylonians arrived at this rule. Almost all cuneiform texts found so far provide only problems with solutions laid out in the form of recipes, with no indication as to how they were found.

Despite high level development of algebra in Babylon, the cuneiform texts lack the concept of a negative number and general methods solving quadratic equations.

1.2 How Diophantus composed and solved quadratic equations.

Diophantus' Arithmetic does not contain a systematic presentation of algebra, but it contains a systematic series of problems, accompanied by explanations and solved by constructing equations of various degrees.

When composing equations, Diophantus skillfully selects unknowns to simplify the solution.

Here, for example, is one of his tasks.

Problem 11.“Find two numbers, knowing that their sum is 20 and their product is 96”

Diophantus reasons as follows: from the conditions of the problem it follows that the required numbers are not equal, since if they were equal, then their product would not be equal to 96, but to 100. Thus, one of them will be more than half of their sum, i.e. . 10 + x, the other is less, i.e. 10's. The difference between them 2x.

Hence the equation:

(10 + x)(10 - x) = 96

100's 2 = 96

X 2 - 4 = 0 (1)

From here x = 2. One of the required numbers is equal to 12 , other 8 . Solution x = -2 for Diophantus does not exist, since Greek mathematics knew only positive numbers.

If we solve this problem by choosing one of the required numbers as the unknown, then we will come to a solution to the equation

y(20 - y) = 96,

at2 - 20у + 96 = 0. (2)

It is clear that by choosing the half-difference of the required numbers as the unknown, Diophantus simplifies the solution; he manages to reduce the problem to solving an incomplete quadratic equation (1).

1.3 Quadratic Equations in India

Problems on quadratic equations are found already in the astronomical treatise “Aryabhattiam”, compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (7th century), outlined a general rule for solving quadratic equations reduced to a single canonical form:

Oh2 + bx = c, a > 0. (1)

In equation (1), the coefficients, except A, can also be negative. Brahmagupta's rule is essentially the same as ours.

In ancient India, public competitions in solving difficult problems were common. One of the old Indian books says the following about such competitions: “As the sun eclipses the stars with its brilliance, so learned man eclipse the glory of another in popular assemblies by proposing and solving algebraic problems.” Problems were often presented in poetic form.

This is one of the problems of the famous Indian mathematician of the 12th century. Bhaskars.

Problem 13.

“A flock of frisky monkeys, and twelve along the vines...

The authorities, having eaten, had fun. They started jumping, hanging...

There are them in the square, part eight. How many monkeys were there?

I was having fun in the clearing. Tell me, in this pack?

Bhaskara's solution indicates that he knew that the roots of quadratic equations are two-valued (Fig. 3).

The equation corresponding to problem 13 is:

(x/8) 2 + 12 = x

Bhaskara writes under the guise:

X2 - 64x = -768

and to complement left side of this equation to the square, adds to both sides 32 2 , then getting:

X2 - 64x + 322 = -768 + 1024,

(x - 32)2 = 256,

x - 32 = ± 16,

X1 = 16, x2 = 48.

1.4 Quadratic equations in al - Khorezmi

In the algebraic treatise of al-Khorezmi, a classification of linear and quadratic equations is given. The author counts 6 types of equations, expressing them as follows:

1) “Squares are equal to roots,” i.e. Oh2 + c =bX.

2) “Squares are equal to numbers”, i.e. Oh2 = s.

3) “The roots are equal to the number,” i.e. ah = s.

4) “Squares and numbers are equal to roots,” i.e. Oh2 + c =bX.

5) “Squares and roots are equal to numbers”, i.e. Oh2 + bx= s.

6) “Roots and numbers are equal to squares,” i.e.bx+ c = ah2 .

al-Khorezmi, like all mathematicians before the 17th century, does not take into account the zero solution, probably because in specific practical problems it does not matter. When solving complete quadratic equations, al-Khorezmi sets out the rules for solving them using particular numerical examples, and then geometric proofs.

Problem 14.“The square and the number 21 are equal to 10 roots. Find the root" (assuming the root of the equation x2 + 21 = 10x).

The author's solution goes something like this: divide the number of roots in half, you get 5, multiply 5 by itself, subtract 21 from the product, what remains is 4. Take the root from 4, you get 2. Subtract 2 from 5, you get 3, this will be the desired root. Or add 2 to 5, which gives 7, this is also a root.

The treatise of al-Khorezmi is the first book that has come down to us, which systematically sets out the classification of quadratic equations and gives formulas for their solution.

1.5 Quadratic equations in EuropeXIII- XVIIbb

Formulas for solving quadratic equations along the lines of al-Khwarizmi in Europe were first set forth in the Book of Abacus, written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both Islamic countries and Ancient Greece, is distinguished by both completeness and clarity of presentation. The author independently developed some new algebraic examples of solving problems and was the first in Europe to approach the introduction of negative numbers. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many problems from the Book of Abacus were used in almost all European textbooks of the 16th - 17th centuries. and partly XVIII.

PAGE_BREAK--

The general rule for solving quadratic equations reduced to a single canonical form:

X2 + bx= c,

for all possible combinations of coefficient signs b, With was formulated in Europe only in 1544 by M. Stiefel.

Derivation of the formula for solving a quadratic equation in general view Viet has it, but Viet recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. In addition to positive ones, negative roots are also taken into account. Only in the 17th century. Thanks to the work of Girard, Descartes, Newton and other scientists, the method of solving quadratic equations takes on a modern form.

1.6 About Vieta's theorem

The theorem expressing the relationship between the coefficients of a quadratic equation and its roots, named after Vieta, was formulated by him for the first time in 1591 as follows: “If B+ D, multiplied by A- A2 , equals BD, That A equals IN and equal D».

To understand Vieta, we should remember that A, like any vowel letter, meant the unknown (our X), vowels IN,D- coefficients for the unknown. In the language of modern algebra, the above Vieta formulation means: if there is

(a +b)x - x2 = ab,

X2 - (a +b)x + ab= 0,

X1 = a, x2 = b.

Expressing the relationship between the roots and coefficients of equations with general formulas written using symbols, Viète established uniformity in the methods of solving equations. However, the symbolism of Viet is still far from modern look. He did not recognize negative numbers and therefore, when solving equations, he considered only cases where all the roots were positive.

2. Methods for solving quadratic equations

Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve quadratic equations from school (8th grade) until graduation.

In the school mathematics course, formulas for the roots of quadratic equations are studied, with the help of which you can solve any quadratic equations. However, there are other ways to solve quadratic equations that allow you to solve many equations very quickly and efficiently. There are ten ways to solve quadratic equations. In my work, I analyzed each of them in detail.

1. METHOD : Factoring the left side of the equation.

Let's solve the equation

X2 + 10x - 24 = 0.

Let's factorize the left side:

X2 + 10x - 24 = x2 + 12x - 2x - 24 = x(x + 12) - 2(x + 12) = (x + 12)(x - 2).

Therefore, the equation can be rewritten as follows:

(x + 12)(x - 2) = 0

Since the product is zero, then at least one of its factors is zero. Therefore, the left side of the equation becomes zero at x = 2, and also when x = - 12. This means that the number 2 And - 12 are the roots of the equation X2 + 10x - 24 = 0.

2. METHOD : Method for selecting a complete square.

Let's solve the equation X2 + 6x - 7 = 0.

Select a complete square on the left side.

To do this, we write the expression x2 + 6x in the following form:

X2 + 6x = x2 + 2 x 3.

In the resulting expression, the first term is the square of the number x, and the second is the double product of x by 3. Therefore, to get a complete square, you need to add 32, since

x2 + 2 x 3 + 32 = (x + 3)2 .

Let us now transform the left side of the equation

X2 + 6x - 7 = 0,

adding to it and subtracting 32. We have:

X2 + 6x - 7 = x2 + 2 x 3 + 32 - 3 2 - 7 = (x + 3)2 - 9 - 7 = (x + 3)2 - 16.

Thus, this equation can be written as follows:

(x + 3)2 - 16 =0, (x + 3)2 = 16.

Hence, x + 3 - 4 = 0, x1 = 1, or x + 3 = -4, x2 = -7.

3. METHOD :Solving quadratic equations using the formula.

Let's multiply both sides of the equation

Oh2 + bx + c = 0, a ≠ 0

on 4a and sequentially we have:

4a2 X2 + 4abx + 4ac = 0,

((2ah)2 + 2ahb+ b2 ) - b2 + 4 ac= 0,

(2ax + b)2 = b2 - 4ac,

2ax + b = ± √ b2 - 4ac,

2ax = - b ± √ b2 - 4ac,

Examples.

A) Let's solve the equation: 4x2 + 7x + 3 = 0.

a = 4,b= 7, s = 3,D= b2 - 4 ac= 7 2 - 4 4 3 = 49 - 48 = 1,

D> 0, two different roots;

Thus, in the case of a positive discriminant, i.e. at

b2 - 4 ac>0 , the equation Oh2 + bx + c = 0 has two different roots.

b) Let's solve the equation: 4x2 - 4x + 1 = 0,

a = 4,b= - 4, s = 1,D= b2 - 4 ac= (-4) 2 - 4 4 1= 16 - 16 = 0,

D= 0, one root;

So, if the discriminant is zero, i.e. b2 - 4 ac= 0 , then the equation

Oh2 + bx + c = 0 has a single root

V) Let's solve the equation: 2x2 + 3x + 4 = 0,

a = 2,b= 3, c = 4,D= b2 - 4 ac= 3 2 - 4 2 4 = 9 - 32 = - 13, D< 0.

Continuation
--PAGE_BREAK--

This equation has no roots.

So, if the discriminant is negative, i.e. b2 - 4 ac< 0 ,

the equation Oh2 + bx + c = 0 has no roots.

Formula (1) of the roots of a quadratic equation Oh2 + bx + c = 0 allows you to find roots any quadratic equation (if any), including reduced and incomplete. Formula (1) is expressed verbally as follows: the roots of a quadratic equation are equal to a fraction whose numerator is equal to the second coefficient taken with the opposite sign, plus minus the square root of the square of this coefficient without quadruple the product of the first coefficient by the free term, and the denominator is double the first coefficient.

4. METHOD: Solving equations using Vieta's theorem.

As is known, the reduced quadratic equation has the form

X2 + px+ c= 0. (1)

Its roots satisfy Vieta’s theorem, which, when a =1 looks like

/>x1 x2 = q,

x1 + x2 = - p

From this we can draw the following conclusions (from the coefficients p and q we can predict the signs of the roots).

a) If the half-member q given equation (1) is positive ( q> 0 ), then the equation has two roots of equal sign and this depends on the second coefficient p. If R< 0 , then both roots are negative if R< 0 , then both roots are positive.

For example,

x2 – 3 x+ 2 = 0; x1 = 2 And x2 = 1, because q= 2 > 0 And p= - 3 < 0;

x2 + 8 x+ 7 = 0; x1 = - 7 And x2 = - 1, because q= 7 > 0 And p= 8 > 0.

b) If a free member q given equation (1) is negative ( q< 0 ), then the equation has two roots of different sign, and the larger root will be positive if p< 0 , or negative if p> 0 .

For example,

x2 + 4 x– 5 = 0; x1 = - 5 And x2 = 1, because q= - 5 < 0 And p= 4 > 0;

x2 – 8 x– 9 = 0; x1 = 9 And x2 = - 1, because q= - 9 < 0 And p= - 8 < 0.

5. METHOD: Solving equations using the "throw" method.

Consider the quadratic equation

Oh2 + bx + c = 0, Where a ≠ 0.

Multiplying both sides by a, we obtain the equation

A2 X2 + abx + ac = 0.

Let ah = y, where x = y/a; then we come to the equation

at2 + by+ ac = 0,

is equivalent to this. Its roots at1 And at 2 can be found using Vieta’s theorem.

Finally we get

X1 = y1 /A And X1 = y2 /A.

With this method the coefficient A multiplied by the free term, as if “thrown” to it, which is why it is called transfer method. This method is used when the roots of the equation can be easily found using Vieta's theorem and, most importantly, when the discriminant is an exact square.

Example.

Let's solve the equation 2x2 – 11x + 15 = 0.

Solution. Let’s “throw” coefficient 2 to the free term, and as a result we get the equation

at2 – 11у + 30 = 0.

According to Vieta's theorem

/>/>/>/>/>at1 = 5 x1 = 5/2 x1 = 2,5

at2 = 6 x2 = 6/2 x2 = 3.

Answer: 2.5; 3.

6. METHOD: Properties of coefficients of a quadratic equation.

A. Let a quadratic equation be given

Oh2 + bx + c = 0, Where a ≠ 0.

1) If, a+b+ c = 0 (i.e. the sum of the coefficients is zero), then x1 = 1,

X2 = s/a.

Proof. Dividing both sides of the equation by a ≠ 0, we obtain the reduced quadratic equation

x2 + b/ a x+ c/ a= 0.

/>According to Vieta’s theorem

x1 + x2 = - b/ a,

x1 x2 = 1 c/ a.

By condition A -b+ c = 0, where b= a + c. Thus,

/>x1 +x2 = - A+ b/a= -1 – c/a,

x1 x2 = - 1 (- c/a),

those. X1 = -1 And X2 = c/ a, which we needed to prove.

Examples.

Let's solve the equation 345x2 – 137x – 208 = 0.

Solution. Because a +b+ c = 0 (345 – 137 – 208 = 0), That

X1 = 1, x2 = c/ a= -208/345.

Answer: 1; -208/345.

2) Solve the equation 132x2 – 247x + 115 = 0.

Solution. Because a +b+ c = 0 (132 – 247 + 115 = 0), That

X1 = 1, x2 = c/ a= 115/132.

Answer: 1; 115/132.

B. If the second coefficient b= 2 k is an even number, then the root formula

Continuation
--PAGE_BREAK--

Example.

Let's solve the equation 3x2 - 14x + 16 = 0.

Solution. We have: a = 3,b= - 14, s = 16,k= - 7 ;

D= k2 – ac= (- 7) 2 – 3 16 = 49 – 48 = 1, D> 0, two different roots;

Answer: 2; 8/3

IN. Reduced equation

X2 + px +q= 0

coincides with a general equation in which a = 1, b= p And c =q. Therefore, for the reduced quadratic equation, the root formula is

takes the form:

Formula (3) is especially convenient to use when R- even number.

Example. Let's solve the equation X2 – 14x – 15 = 0.

Solution. We have: X1,2 =7±

Answer: x1 = 15; X2 = -1.

7. METHOD: Graphical solution of a quadratic equation.

If in Eq.

X2 + px+ q= 0

move the second and third terms to the right side, we get

X2 = - px- q.

Let's build graphs of the dependence y = x2 and y = - px- q.

The graph of the first dependence is a parabola passing through the origin. Second dependency graph -

straight (Fig. 1). The following cases are possible:

A straight line and a parabola can intersect at two points, the abscissas of the intersection points are the roots of the quadratic equation;

A straight line and a parabola can touch (only one common point), i.e. the equation has one solution;

A straight line and a parabola do not have common points, i.e. a quadratic equation has no roots.

Examples.

1) Let's solve the equation graphically X2 - 3x - 4 = 0(Fig. 2).

Solution. Let's write the equation in the form X2 = 3x + 4.

Let's build a parabola y = x2 and direct y = 3x + 4. Direct

y = 3x + 4 can be built from two points M (0; 4) And

N(3; 13) . A straight line and a parabola intersect at two points

A And IN with abscissas X1 = - 1 And X2 = 4 . Answer : X1 = - 1;

X2 = 4.

2) Let's solve the equation graphically (Fig. 3) X2 - 2x + 1 = 0.

Solution. Let's write the equation in the form X2 = 2x - 1.

Let's build a parabola y = x2 and direct y = 2x - 1.

Direct y = 2x - 1 build from two points M (0; - 1)

And N(1/2; 0) . A straight line and a parabola intersect at a point A With

abscissa x = 1. Answer: x = 1.

3) Let's solve the equation graphically X2 - 2x + 5 = 0(Fig. 4).

Solution. Let's write the equation in the form X2 = 5x - 5. Let's build a parabola y = x2 and direct y = 2x - 5. Direct y = 2x - 5 Let's build from two points M(0; - 5) and N(2.5; 0). A straight line and a parabola do not have intersection points, i.e. This equation has no roots.

Answer. The equation X2 - 2x + 5 = 0 has no roots.

8. METHOD: Solving quadratic equations using compass and ruler.

The graphical method of solving quadratic equations using a parabola is inconvenient. If you build a parabola point by point, it takes a lot of time, and the degree of accuracy of the results obtained is low.

I propose the following method for finding the roots of a quadratic equation Oh2 + bx + c = 0 using a compass and ruler (Fig. 5).

Let us assume that the desired circle intersects the axis

abscissa in points B(x1 ; 0) And D(X2 ; 0), Where X1 And X2 - roots of the equation Oh2 + bx + c = 0, and passes through the points

A(0; 1) And C(0;c/ a) on the ordinate axis. Then, by the secant theorem, we have O.B. O.D.= O.A. O.C., where O.C.= O.B. O.D./ O.A.= x1 X2 / 1 = c/ a.

The center of the circle is at the point of intersection of the perpendiculars SF And S.K., restored in the middles of the chords A.C. And BD, That's why

1) construct points (center of the circle) and A(0; 1) ;

2) draw a circle with radius S.A.;

3) abscissa of the points of intersection of this circle with the axis Oh are the roots of the original quadratic equation.

In this case, three cases are possible.

1) The radius of the circle is greater than the ordinate of the center (AS> S.K., orR> a+ c/2 a) , the circle intersects the Ox axis at two points (Fig. 6, a) B(x1 ; 0) And D(X2 ; 0) , Where X1 And X2 - roots of quadratic equation Oh2 + bx + c = 0.

2) The radius of the circle is equal to the ordinate of the center (AS= S.B., orR= a+ c/2 a) , the circle touches the Ox axis (Fig. 6, b) at the point B(x1 ; 0) , where x1 is the root of the quadratic equation.

Continuation
--PAGE_BREAK--

3) The radius of the circle is less than the ordinate of the center; the circle has no common points with the abscissa axis (Fig. 6, c), in this case the equation has no solution.

Example.

Let's solve the equation X2 - 2x - 3 = 0(Fig. 7).

Solution. Let's determine the coordinates of the center point of the circle using the formulas:

Let's draw a circle of radius SA, where A (0; 1).

Answer:X1 = - 1; X2 = 3.

9. METHOD: Solving quadratic equations using a nomogram.

This is an old and undeservedly forgotten method of solving quadratic equations, placed on p. 83 (see Bradis V.M. Four-digit mathematical tables. - M., Prosveshchenie, 1990).

Table XXII. Nomogram for solving the equation z2 + pz+ q= 0 . This nomogram allows, without solving a quadratic equation, to determine the roots of the equation using its coefficients.

The curvilinear scale of the nomogram is built according to the formulas (Fig. 11):

Believing OS = p,ED= q, OE = a(all in cm), from the similarity of triangles SAN And CDF we get the proportion

which, after substitutions and simplifications, yields the equation

z2 + pz+ q= 0,

and the letter z means the mark of any point on a curved scale.

Examples.

1) For the equation z2 - 9 z+ 8 = 0 nomogram gives roots

z1 = 8,0 And z2 = 1,0 (Fig. 12).

2) Using a nomogram, we solve the equation

2 z2 - 9 z+ 2 = 0.

Dividing the coefficients of this equation by 2, we get the equation

z2 - 4,5 z+ 1 = 0.

Nomogram gives roots z1 = 4 And z2 = 0,5.

3) For the equation

z2 - 25 z+ 66 = 0

coefficients p and q are outside the scale, let’s perform the substitution z= 5 t, we get the equation

t2 - 5 t+ 2,64 = 0,

which we solve using a nomogram and get t1 = 0,6 And t2 = 4,4, where z1 = 5 t1 = 3,0 And z2 = 5 t2 = 22,0.

10. METHOD: Geometric method for solving quadratic equations.

In ancient times, when geometry was more developed than algebra, quadratic equations were solved not algebraically, but geometrically. I will give a famous example from the “Algebra” of al-Khorezmi.

Examples.

1) Let's solve the equation X2 + 10x = 39.

In the original, this problem is formulated as follows: “A square and ten roots are equal to 39” (Fig. 15).

Solution. Consider a square with side x, rectangles are constructed on its sides so that the other side of each of them is 2.5, therefore, the area of each is 2.5x. The resulting figure is then supplemented to a new square ABCD, building four equal squares in the corners, the side of each of them is 2.5, and the area is 6.25.

Square S square ABCD can be represented as the sum of areas: the original square X2 , four rectangles (4 2.5x = 10x) and four attached squares (6,25 4 = 25) , i.e. S= X2 + 10x + 25. Replacing

X2 + 10x number 39 , we get that S= 39 + 25 = 64 , which means that the side of the square ABCD, i.e. line segment AB = 8. For the required side X we get the original square

2) But, for example, how the ancient Greeks solved the equation at2 + 6у - 16 = 0.

Solution shown in Fig. 16, where

at2 + 6y = 16, or y2 + 6y + 9 = 16 + 9.

Solution. Expressions at2 + 6у + 9 And 16 + 9 geometrically represent the same square, and the original equation at2 + 6у - 16 + 9 - 9 = 0- the same equation. From where we get that y + 3 = ± 5, or at1 = 2, y2 = - 8 (Fig. 16).

3) Solve the geometric equation at2 - 6у - 16 = 0.

Transforming the equation, we get

at2 - 6y = 16.

In Fig. 17 find “images” of the expression at2 - 6u, those. from the area of a square with side y, subtract the area of a square with side equal to 3 . This means that if to the expression at2 - 6у add 9 , then we get the area of a square with side y - 3. Replacing the expression at2 - 6у its equal number 16,

we get: (y - 3)2 = 16 + 9, those. y - 3 = ± √25, or y - 3 = ± 5, where at1 = 8 And at2 = - 2.

Conclusion

Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities.

However, the significance of quadratic equations lies not only in the elegance and brevity of solving problems, although this is very important. It is equally important that as a result of the use of quadratic equations in solving problems, new details are often discovered, interesting generalizations can be made and clarifications can be made, which are suggested by the analysis of the resulting formulas and relationships.

I would also like to note that the topic presented in this work has not yet been studied much at all, it is simply not being studied, so it is fraught with a lot of hidden and unknown things, which provides an excellent opportunity for further work on it.

Here I dwelled on the issue of solving quadratic equations, and what,

if there are other ways to solve them?! Again, find beautiful patterns, some facts, clarifications, make generalizations, discover more and more new things. But these are questions for future work.

To summarize, we can conclude: quadratic equations play a huge role in the development of mathematics. We all know how to solve quadratic equations from school (8th grade) until graduation. This knowledge can be useful to us throughout our lives.

Since these methods for solving quadratic equations are easy to use, they should certainly be of interest to students who are interested in mathematics. My work makes it possible to look differently at the tasks that mathematics poses to us.

Literature:

1. Alimov Sh.A., Ilyin V.A. and others. Algebra, 6-8. Trial textbook for 6-8 grade high school. - M., Education, 1981.

2. Bradis V.M. Four-digit math tables for high school. Ed. 57th. - M., Education, 1990. P. 83.

3. Kruzhepov A.K., Rubanov A.T. Problem book on algebra and elementary functions. Textbook for secondary special education educational institutions. - M., higher school, 1969.

4. Okunev A.K. Quadratic functions, equations and inequalities. Teacher's manual. - M., Education, 1972.

5. Presman A.A. Solving a quadratic equation using a compass and ruler. - M., Kvant, No. 4/72. P. 34.

6. Solomnik V.S., Milov P.I. Collection of questions and problems in mathematics. Ed. - 4th, additional - M., Higher School, 1973.

7. Khudobin A.I. Collection of problems on algebra and elementary functions. Teacher's manual. Ed. 2nd. - M., Education, 1970.

MINISTRY OF EDUCATION AND SCIENCE OF THE RF

Bryansk region Zhukovsky district

Municipal educational institution Rzhanitsky secondary school

RESEARCH

SOLUTION WAYS

Pavlikov Dmitry, 9th grade.

Head: Prikhodko Yuri
Vladimirovich,

mathematic teacher.

BRYANSK, 2009

I. History of the development of quadratic equations ……………………….2

1. Quadratic equations in Ancient Babylon………………………..2

2. How Diophantus composed and solved quadratic equations…………...2

3. Quadratic equations in India……………………………………...3

4. Quadratic equations in al-Khorezmi……………………………4

5. Quadratic equations in Europe XIII - XVII centuries……………….........5

6. About Vieta’s theorem………………………………………………………6

II. Methods for solving quadratic equations ……………………….7

Method…………………………………………………………………………………7

Method……………………………………………………………………...9

Method……………………………………………………………...10

Method……………………………………………………………...12

Method……………………………………………………………...13

Method……………………………………………………………...15

Method……………………………………………………………...16

III. Conclusion…………………………………………………..............18

Literature……………………………………………………………….19

History of the development of quadratic equations.

1. Quadratic equations in Ancient Babylon.

Using modern algebraic notation, we can say that in their cuneiform texts there are, in addition to incomplete ones, such, for example, complete quadratic equations:

X 2 + X = ¾; X 2 - X = 14,5

Despite the high level of development of algebra in Babylon, the cuneiform texts lack the concept of a negative number and general methods for solving quadratic equations.

2. How Diophantus composed and solved quadratic equations.

When composing equations, Diophantus skillfully selects unknowns to simplify the solution.

Here, for example, is one of his tasks.

Problem 11.“Find two numbers, knowing that their sum is 20 and their product is 96”

Hence the equation:

(10 + x)(10 - x) = 96

100's 2 = 96

X 2 - 4 = 0 (1)

From here x = 2. One of the required numbers is equal to 12 , other 8 . Solution x = -2 for Diophantus does not exist, since Greek mathematics knew only positive numbers.

If we solve this problem by choosing one of the required numbers as the unknown, then we will come to a solution to the equation

y(20 - y) = 96,

at 2 - 20у + 96 = 0. (2)

3. Quadratic equations in India.

Oh 2 + bx = c, a 0. (1)

In equation (1), the coefficients, except A, can also be negative. Brahmagupta's rule is essentially the same as ours.

In ancient India, public competitions in solving difficult problems were common. One of the old Indian books says the following about such competitions: “As the sun outshines the stars with its brilliance, so a learned man will outshine the glory of another in public assemblies, proposing and solving algebraic problems.” Problems were often presented in poetic form.

This is one of the problems of the famous Indian mathematician of the 12th century. Bhaskars.

Problem 13.

“A flock of frisky monkeys, and twelve along the vines...

The authorities, having eaten, had fun. They started jumping, hanging...

There are them in the square, part eight. How many monkeys were there?

I was having fun in the clearing. Tell me, in this pack?

Bhaskara's solution indicates that he knew that the roots of quadratic equations are two-valued (Fig. 3).

The equation corresponding to problem 13 is:

(x/8) 2 + 12 = x

Bhaskara writes under the guise:

X 2 - 64x = -768

and, to complete the left side of this equation to square, adds to both sides 32 2 , then getting:

X 2 - 64x + 32 2 = -768 + 1024,

(x - 32) 2 = 256,

x - 32 = ± 16,

X 1 = 16, x 2 = 48.

4. Quadratic equations of al-Khorezmi.

In the algebraic treatise of al-Khorezmi, a classification of linear and quadratic equations is given. The author counts 6 types of equations, expressing them as follows:

1) “Squares are equal to roots,” i.e. Oh 2 + c =bX.

2) “Squares are equal to numbers”, i.e. Oh 2 = s.

3) “The roots are equal to the number,” i.e. ah = s.

4) “Squares and numbers are equal to roots,” i.e. Oh 2 + c =bX.

5) “Squares and roots are equal to numbers”, i.e. Oh 2 + bx= s.

6) “Roots and numbers are equal to squares,” i.e.bx+ c = ah 2 .

For al-Khorezmi, who avoided the use of negative numbers, the terms of each of these equations are addends and not subtractables. In this case, equations that do not have positive solutions are obviously not taken into account. The author sets out methods for solving these equations using the techniques of al-jabr and al-muqabala. His decisions, of course, do not completely coincide with ours. Not to mention that it is purely rhetorical, it should be noted, for example, that when solving an incomplete quadratic equation of the first type, al-Khorezmi, like all mathematicians before the 17th century, takes into account the zero solution, probably because in specific practical it doesn't matter in tasks. When solving complete quadratic equations, al-Khorezmi sets out the rules for solving them using particular numerical examples, and then geometric proofs.

Here's an example:

Problem 14.“The square and the number 21 are equal to 10 roots. Find the root"

(assuming the root of the equation x 2 + 21 = 10x).

The treatise of al-Khorezmi is the first book that has come down to us, which systematically sets out the classification of quadratic equations and gives formulas for their solution.

5. Quadratic equations in EuropeXIII - XVIIcenturies

Formulas for solving quadratic equations along the lines of al-Khwarizmi in Europe were first set forth in the Book of Abacus, written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both from the countries of Islam and from ancient Greece, is distinguished by its completeness and clarity of presentation. The author independently developed some new algebraic examples of solving problems and was the first in Europe to approach the introduction of negative numbers. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many problems from the Book of Abacus were used in almost all European textbooks of the 16th - 17th centuries. and partly XVIII.

The general rule for solving quadratic equations reduced to a single canonical form:

X 2 + bx= c,

for all possible combinations of coefficient signs b, With was formulated in Europe only in 1544 by M. Stiefel.

The derivation of the formula for solving a quadratic equation in general form is available from Viète, but Viète recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. In addition to positive ones, negative roots are also taken into account. Only in the 17th century. Thanks to the work of Girard, Descartes, Newton and other scientists, the method of solving quadratic equations takes on a modern form.

6. About Vieta's theorem.

The theorem expressing the relationship between the coefficients of a quadratic equation and its roots, named after Vieta, was formulated by him for the first time in 1591 as follows: “If B + D, multiplied by A - A 2 , equals BD, That A equals IN and equal D».

(a +b)x - x 2 = ab,

X 2 - (a +b)x + ab = 0,

X 1 = a, x 2 = b.

Expressing the relationship between the roots and coefficients of equations with general formulas written using symbols, Viète established uniformity in the methods of solving equations. However, the symbolism of Viet is still far from its modern form. He did not recognize negative numbers and therefore, when solving equations, he considered only cases where all the roots were positive.

So: Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve quadratic equations from school (8th grade) until graduation.

1. METHOD : Factoring the left side of the equation.

Let's solve the equation X 2 + 10x - 24 = 0. Let's factorize the left side:

X 2 + 10x - 24 = x 2 + 12x - 2x - 24 = x(x + 12) - 2(x + 12) = (x + 12)(x - 2).

Therefore, the equation can be rewritten as follows:

(x + 12)(x - 2) = 0

2. METHOD : Method for selecting a complete square.

Let's solve the equation X 2 + 6x - 7 = 0. Select a complete square on the left side.

To do this, we write the expression x 2 + 6x in the following form:

X 2 + 6x = x 2 + 2 X 3.

In the resulting expression, the first term is the square of the number x, and the second is the double product of x by 3. Therefore, to get a complete square, you need to add 3 2, since

x 2 + 2 X 3 + 3 2 = (x + 3) 2 .

Let us now transform the left side of the equation

X 2 + 6x - 7 = 0,

adding to it and subtracting 3 2. We have:

X 2 + 6x - 7 = x 2 + 2 X 3 + 3 2 - 3 2 - 7 = (x + 3) 2 - 9 - 7 = (x + 3) 2 - 16.

Thus, this equation can be written as follows:

(x + 3) 2 - 16 =0, (x + 3) 2 = 16.

Hence, x + 3 - 4 = 0, x 1 = 1, or x + 3 = -4, x 2 = -7.

3. METHOD :Solving quadratic equations using the formula.

Let's multiply both sides of the equation

Oh 2 + bx + c = 0, a ≠ 0

on 4a and sequentially we have:

4a 2 X 2 + 4abx + 4ac = 0,

((2ah) 2 + 2ah b + b 2 ) - b 2 + 4 ac = 0,

(2ax + b) 2 = b 2 - 4ac,

2ax + b = ± √ b 2 - 4ac,

2ax = - b ± √ b 2 - 4ac,

Examples.

A) Let's solve the equation: 4x 2 + 7x + 3 = 0.

a = 4,b= 7, s = 3,D = b 2 - 4 ac = 7 2 - 4 4 3 = 49 - 48 = 1,

D 0, two different roots;

Thus, in the case of a positive discriminant, i.e. at

b 2 - 4 ac 0 , the equation Oh 2 + bx + c = 0 has two different roots.

b) Let's solve the equation: 4x 2 - 4x + 1 = 0,

a = 4,b= - 4, s = 1,D = b 2 - 4 ac = (-4) 2 - 4 4 1= 16 - 16 = 0,

D = 0, one root;

So, if the discriminant is zero, i.e. b 2 - 4 ac = 0 , then the equation

Oh 2 + bx + c = 0 has a single root

V) Let's solve the equation: 2x 2 + 3x + 4 = 0,

a = 2,b= 3, c = 4,D = b 2 - 4 ac = 3 2 - 4 2 4 = 9 - 32 = - 13 , D

This equation has no roots.

So, if the discriminant is negative, i.e. b 2 - 4 ac, the equation

Oh 2 + bx + c = 0 has no roots.

Formula (1) of the roots of a quadratic equation Oh 2 + bx + c = 0 allows you to find roots any quadratic equation (if any), including reduced and incomplete. Formula (1) is expressed verbally as follows: the roots of a quadratic equation are equal to a fraction whose numerator is equal to the second coefficient taken with the opposite sign, plus minus the square root of the square of this coefficient without quadruple the product of the first coefficient by the free term, and the denominator is double the first coefficient.

4. METHOD: Solving equations using Vieta's theorem.

As is known, the reduced quadratic equation has the form

X 2 + px + c = 0. (1)

Its roots satisfy Vieta’s theorem, which, when a =1 looks like

x 1 x 2 = q,

x 1 + x 2 = - p

From this we can draw the following conclusions (from the coefficients p and q we can predict the signs of the roots).

a) If the half-member q given equation (1) is positive ( q 0 ), then the equation has two roots of equal sign and this depends on the second coefficient p. If p, then both roots are negative if p, then both roots are positive.

For example,

x 2 – 3 x + 2 = 0; x 1 = 2 And x 2 = 1, because q = 2 0 And p = - 3

x 2 + 8 x + 7 = 0; x 1 = - 7 And x 2 = - 1, because q = 7 0 And p= 8 0.

b) If a free member q given equation (1) is negative ( q), then the equation has two roots of different sign, and the larger root will be positive if p, or negative if p 0 .

For example,

x 2 + 4 x – 5 = 0; x 1 = - 5 And x 2 = 1, because q= - 5 and p = 4 0;

x 2 – 8 x – 9 = 0; x 1 = 9 And x 2 = - 1, because q= - 9 and p = - 8

5. METHOD: Solving equations using the "throw" method.

Consider the quadratic equation

Oh 2 + bx + c = 0, Where a ≠ 0.

Multiplying both sides by a, we obtain the equation

A 2 X 2 + abx + ac = 0.

Let ah = y, where x = y/a; then we come to the equation

at 2 + by+ ac = 0,

is equivalent to this. Its roots at 1 And at 2 can be found using Vieta’s theorem.

Finally we get X 1 = y 1 /A And X 1 = y 2 /A. With this method the coefficient A multiplied by the free term, as if “thrown” to it, which is why it is called transfer method. This method is used when the roots of the equation can be easily found using Vieta's theorem and, most importantly, when the discriminant is an exact square.

Example.

Let's solve the equation 2x 2 – 11x + 15 = 0.

Solution. Let’s “throw” coefficient 2 to the free term, and as a result we get the equation

at 2 – 11у + 30 = 0.

According to Vieta's theorem

at1 = 5 x 1 = 5/2 x 1 = 2,5

at 2 = 6 x 2 = 6/2 x 2 = 3.

Answer: 2.5; 3.

6. METHOD: Properties of coefficients of a quadratic equation.

A. Let a quadratic equation be given Oh 2 + bx + c = 0, Where a ≠ 0.

1) If, a+b+ c = 0 (i.e. the sum of the coefficients is zero), then x 1 = 1,

X 2 = s/a.

Proof. Dividing both sides of the equation by a ≠ 0, we obtain the reduced quadratic equation

x 2 + b/ a x + c/ a = 0.

According to Vieta's theorem

x 1 + x 2 = - b/ a,

x 1 x 2 = 1 c/ a.

By condition A -b+ c = 0, where b= a + c. Thus,

x 1 + x 2 = - a +b/ a= -1 – c/ a,

x 1 x 2 = - 1 (- c/ a),

those. X 1 = -1 And X 2 = c/ a, which we needed to prove.

Examples.

Let's solve the equation 345x 2 – 137x – 208 = 0.

Solution. Because a +b+ c = 0 (345 – 137 – 208 = 0), That

X 1 = 1, x 2 = c/ a = -208/345.

Answer: 1; -208/345.

2) Solve the equation 132x 2 – 247x + 115 = 0.

Solution. Because a +b+ c = 0 (132 – 247 + 115 = 0), That

X 1 = 1, x 2 = c/ a = 115/132.

Answer: 1; 115/132.

B. If the second coefficient b = 2 k is an even number, then the root formula

Example.

Let's solve the equation 3x2 - 14x + 16 = 0.

Solution. We have: a = 3,b= - 14, s = 16,k = - 7 ;

D = k 2 – ac = (- 7) 2 – 3 16 = 49 – 48 = 1, D 0, two different roots;

Answer: 2; 8/3

IN. Reduced equation

X 2 + px +q= 0

coincides with a general equation in which a = 1, b= p And c =q. Therefore, for the reduced quadratic equation, the root formula is

takes the form:

Formula (3) is especially convenient to use when R- even number.

Example. Let's solve the equation X 2 – 14x – 15 = 0.

Solution. We have: X 1,2 =7±

Answer: x 1 = 15; X 2 = -1.

7. METHOD: Graphical solution of a quadratic equation.

E if in the equation

X 2 + px + q = 0

move the second and third terms to the right side, we get

X 2 = - px - q.

Let's build graphs of the dependence y = x 2 and y = - px - q.

The graph of the first dependence is a parabola passing through the origin. Second dependency graph -

straight (Fig. 1). The following cases are possible:

A straight line and a parabola can intersect at two points,

the abscissas of the intersection points are the roots of the quadratic equation;

A straight line and a parabola can touch (only one common point), i.e. the equation has one solution;

A straight line and a parabola do not have common points, i.e. a quadratic equation has no roots.

Examples.

1) Let's solve the equation graphically X 2 - 3x - 4 = 0(Fig. 2).

Solution. Let's write the equation in the form X 2 = 3x + 4.

Let's build a parabola y = x 2 and direct y = 3x + 4. Direct

y = 3x + 4 can be built from two points M (0; 4) And

N (3; 13) . A straight line and a parabola intersect at two points

A And IN with abscissas X 1 = - 1 And X 2 = 4 . Answer : X 1 = - 1;

X 2 = 4.

2) Let's solve the equation graphically (Fig. 3) X 2 - 2x + 1 = 0.

Solution. Let's write the equation in the form X 2 = 2x - 1.

Let's build a parabola y = x 2 and direct y = 2x - 1.

Direct y = 2x - 1 build from two points M (0; - 1)

And N(1/2; 0) . A straight line and a parabola intersect at a point A With

abscissa x = 1. Answer: x = 1.

3) Let's solve the equation graphically X 2 - 2x + 5 = 0(Fig. 4).

Solution. Let's write the equation in the form X 2 = 5x - 5. Let's build a parabola y = x 2 and direct y = 2x - 5. Direct y = 2x - 5 Let's build from two points M(0; - 5) and N(2.5; 0). A straight line and a parabola do not have intersection points, i.e. This equation has no roots.

Answer. The equation X 2 - 2x + 5 = 0 has no roots.

8. METHOD: Solving quadratic equations using compasses and

rulers.

I propose the following method for finding the roots of a quadratic equation Oh 2 + bx + c = 0 using a compass and ruler (Fig. 5).

Let us assume that the desired circle intersects the axis

abscissa in points B(x 1 ; 0) And D(X 2 ; 0), Where X 1 And X 2 - roots of the equation Oh 2 + bx + c = 0, and passes through the points

A(0; 1) And C(0;c/ a) on the ordinate axis. Then, by the secant theorem, we have O.B. O.D. = O.A. O.C., where O.C. = O.B. O.D./ O.A.= x 1 X 2 / 1 = c/ a.

The center of the circle is at the point of intersection of the perpendiculars SF And S.K., restored in the middles of the chords A.C. And BD, That's why

1) construct points (center of the circle) and A(0; 1) ;

2) draw a circle with radius S.A.;

3) abscissa of the points of intersection of this circle with the axis Oh are the roots of the original quadratic equation.

In this case, three cases are possible.

1) The radius of the circle is greater than the ordinate of the center (AS S.K., orR a + c/2 a) , the circle intersects the Ox axis at two points (Fig. 6, a) B(x 1 ; 0) And D(X 2 ; 0) , Where X 1 And X 2 - roots of quadratic equation Oh 2 + bx + c = 0.

2) The radius of the circle is equal to the ordinate of the center (AS = S.B., orR = a + c/2 a) , the circle touches the Ox axis (Fig. 6,b) at the point B(x 1 ; 0) , where x 1 is the root of the quadratic equation.

3) The radius of the circle is less than the ordinate of the center

the circle does not have common points with the abscissa axis (Fig. 6, c), in this case the equation has no solution.

Example.

Let's solve the equation X 2 - 2x - 3 = 0(Fig. 7).

Solution. Let's determine the coordinates of the center point of the circle using the formulas:

Let's draw a circle of radius SA, where A (0; 1).

Answer: X 1 = - 1; X 2 = 3.

9. METHOD: Solving quadratic equations using

nomograms.

This is an old and undeservedly forgotten method of solving quadratic equations,

placed on p. 83 (see Bradis V.M. Four-digit mathematical tables. - M., Education, 1990).

Table XXII. Nomogram for solving the equation z 2 + pz + q = 0 . This nomogram allows, without solving the quadratic equation, by its coefficient

there to determine the roots of the equation.

The curvilinear scale of the nomogram is constructed

according to the formulas (Fig. 11):

Believing OS = p,ED = q, OE = a(all in cm), from

similarity of triangles SAN And CDF we get

proportion

which, after substitutions and simplifications, yields the equation

z 2 + pz + q = 0,

and the letter z means the mark of any point on a curved scale.

Examples.

1) For the equation z 2 - 9 z + 8 = 0 nomogram gives roots z 1 = 8,0 And z 2 = 1,0 (Fig. 12).

2) Let us solve the equation using the nomogram

2 z 2 - 9 z + 2 = 0.

Let's divide the coefficients of this equation by 2,

we get the equation

z 2 - 4,5 z + 1 = 0.

Nomogram gives roots z 1 = 4 And z 2 = 0,5.

3) For the equation

z 2 - 25 z + 66 = 0

coefficients p and q are outside the scale, let’s perform the substitution z = 5 t,

we get the equation

t 2 - 5 t + 2,64 = 0,

which we solve using a nomogram and get t 1 = 0,6 And t 2 = 4,4, where z 1 = 5 t 1 = 3,0 And z 2 = 5 t 2 = 22,0.

10. METHOD: Geometric method for solving squares

equations.

Examples.

1) Let's solve the equation X 2 + 10x = 39.

In the original, this problem is formulated as follows: “A square and ten roots are equal to 39” (Fig. 15).

Square S square ABCD can be represented as the sum of areas: the original square X 2 , four rectangles (4 2.5x = 10x) and four attached squares (6,25 4 = 25) , i.e. S = X 2 + 10x + 25. Replacing

X 2 + 10x number 39 , we get that S = 39 + 25 = 64 , which means that the side of the square ABCD, i.e. line segment AB = 8. For the required side X we get the original square

2) But, for example, how the ancient Greeks solved the equation at 2 + 6у - 16 = 0.

Solution shown in Fig. 16, where

at 2 + 6y = 16, or y 2 + 6y + 9 = 16 + 9.

Solution. Expressions at 2 + 6у + 9 And 16 + 9 geometrically represent

the same square, and the original equation at 2 + 6у - 16 + 9 - 9 = 0- the same equation. From where we get that y + 3 = ± 5, or at 1 = 2, y 2 = - 8 (Fig. 16).

3) Solve the geometric equation at 2 - 6у - 16 = 0.

Transforming the equation, we get

at 2 - 6y = 16.

In Fig. 17 find “images” of the expression at 2 - 6u, those. from the area of a square with side y, subtract the area of a square with side equal to 3 . This means that if to the expression at 2 - 6у add 9 , then we get the area of a square with side y - 3. Replacing the expression at 2 - 6у its equal number 16,

we get: (y - 3) 2 = 16 + 9, those. y - 3 = ± √25, or y - 3 = ± 5, where at 1 = 8 And at 2 = - 2.

Conclusion

Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities.

Here we focused on the issue of solving quadratic equations, but what if there are other ways to solve them?! Again, find beautiful patterns, some facts, clarifications, make generalizations, discover more and more new things. But these are questions for future work.

Since these methods for solving quadratic equations are easy to use, they should certainly be of interest to students who are interested in mathematics. Our work makes it possible to take a different look at the tasks that mathematics poses to us.

Literature:

1. Alimov Sh.A., Ilyin V.A. and others. Algebra, 6-8. Trial textbook for 6-8 grade high school. - M., Education, 1981.

2. Bradis V.M. Four Digit Math Worksheets for Middle School.

Ed. 57th. - M., Education, 1990. P. 83.

3. Kruzhepov A.K., Rubanov A.T. Problem book on algebra and elementary functions. Textbook for secondary specialized educational institutions. - M., higher school, 1969.

4. Okunev A.K. Quadratic functions, equations and inequalities. Teacher's manual. - M., Education, 1972.

5. Presman A.A. Solving a quadratic equation using a compass and ruler. - M., Kvant, No. 4/72. P. 34.

6. Solomnik V.S., Milov P.I. Collection of questions and problems in mathematics. Ed. - 4th, additional - M., Higher School, 1973.

7. Khudobin A.I. Collection of problems on algebra and elementary functions. Teacher's manual. Ed. 2nd. - M., Education, 1970.

Application for management

research work

Supervisor: Prikhodko Yuri Vladimirovich (mathematics teacher)

Suggested topic: "10 ways to solve quadratic equations"

Consultants:

Prikhodko Yuri Vladimirovich (mathematics teacher);

Eroshenkov Dmitry Aleksandrovich (computer science teacher)

Educational field of knowledge, academic subject within which the project work is carried out mathematics

Academic disciplines close to the project topic: mathematics

Training class: 9th grade

Composition of the research group: Kursin Dmitry, Pavlikov Dmitry

Type of project on the student's dominant activity: study of rational ways to solve quadratic equations

Type of project by duration: long-term

Type of education: elective course

Necessary equipment: popular science literature related to the consideration of various ways to solve quadratic equations

Intended product of the project: creation of educational and methodological material on the use of rational methods for solving quadratic equations

Kopyevskaya rural secondary school

10 Ways to Solve Quadratic Equations

Head: Patrikeeva Galina Anatolyevna,

mathematic teacher

village Kopevo, 2007

1. History of the development of quadratic equations

1.1 Quadratic equations in Ancient Babylon

1.2 How Diophantus composed and solved quadratic equations

1.3 Quadratic equations in India

1.4 Quadratic equations by al-Khorezmi

1.5 Quadratic equations in Europe XIII - XVII centuries

1.6 About Vieta's theorem

2. Methods for solving quadratic equations

Conclusion

Literature

1. History of the development of quadratic equations

1.1 Quadratic equations in Ancient Babylon

Using modern algebraic notation, we can say that in their cuneiform texts there are, in addition to incomplete ones, such, for example, complete quadratic equations:

X 2 + X = ¾; X 2 - X = 14,5

Despite the high level of development of algebra in Babylon, the cuneiform texts lack the concept of a negative number and general methods for solving quadratic equations.

1.2 How Diophantus composed and solved quadratic equations.

When composing equations, Diophantus skillfully selects unknowns to simplify the solution.

Here, for example, is one of his tasks.

Problem 11.“Find two numbers, knowing that their sum is 20 and their product is 96”

Hence the equation:

(10 + x)(10 - x) = 96

100 - x 2 = 96

x 2 - 4 = 0 (1)

From here x = 2. One of the required numbers is equal to 12 , other 8 . Solution x = -2 for Diophantus does not exist, since Greek mathematics knew only positive numbers.

If we solve this problem by choosing one of the required numbers as the unknown, then we will come to a solution to the equation

y(20 - y) = 96,

y 2 - 20y + 96 = 0. (2)

1.3 Quadratic Equations in India

ah 2 +bx = c, a > 0. (1)

In equation (1), the coefficients, except A, can also be negative. Brahmagupta's rule is essentially the same as ours.

This is one of the problems of the famous Indian mathematician of the 12th century. Bhaskars.

Problem 13.

“A flock of frisky monkeys, and twelve along the vines...

The authorities, having eaten, had fun. They started jumping, hanging...

There are them in the square, part eight. How many monkeys were there?

I was having fun in the clearing. Tell me, in this pack?

Bhaskara's solution indicates that he knew that the roots of quadratic equations are two-valued (Fig. 3).

The equation corresponding to problem 13 is:

(x/8) 2 + 12 = x

Bhaskara writes under the guise:

x 2 - 64x = -768

and, to complete the left side of this equation to square, adds to both sides 32 2 , then getting:

x 2 - 64x + 32 2 = -768 + 1024,

(x - 32) 2 = 256,

x - 32 = ± 16,

x 1 = 16, x 2 = 48.

1.4 Quadratic equations in al - Khorezmi

In the algebraic treatise of al-Khorezmi, a classification of linear and quadratic equations is given. The author counts 6 types of equations, expressing them as follows:

1) “Squares are equal to roots,” i.e. ax 2 + c =bX.

2) “Squares are equal to numbers”, i.e. ax 2 = c.

3) “The roots are equal to the number,” i.e. ah = s.

4) “Squares and numbers are equal to roots,” i.e. ax 2 + c =bX.

5) “Squares and roots are equal to numbers”, i.e. ah 2 +bx= s.

6) “Roots and numbers are equal to squares,” i.e.bx+ c = ax 2 .

Problem 14.“The square and the number 21 are equal to 10 roots. Find the root" (implying the root of the equation x 2 + 21 = 10x).

The treatise of al-Khorezmi is the first book that has come down to us, which systematically sets out the classification of quadratic equations and gives formulas for their solution.

1.5 Quadratic equations in EuropeXIII - XVIIbb

Formulas for solving quadratic equations along the lines of al-Khwarizmi in Europe were first set forth in the Book of Abacus, written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both from the countries of Islam and from ancient Greece, is distinguished by its completeness and clarity of presentation. The author independently developed some new algebraic examples of solving problems and was the first in Europe to approach the introduction of negative numbers. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many problems from the Book of Abacus were used in almost all European textbooks of the 16th - 17th centuries. and partly XVIII.

The general rule for solving quadratic equations reduced to a single canonical form:

x 2 +bx= c,

for all possible combinations of coefficient signs b, With was formulated in Europe only in 1544 by M. Stiefel.

1.6 About Vieta's theorem

The theorem expressing the relationship between the coefficients of a quadratic equation and its roots, named after Vieta, was formulated by him for the first time in 1591 as follows: “If B + D, multiplied by A - A 2 , equals BD, That A equals IN and equal D».

(a +b)x - x 2 =ab,

x 2 - (a +b)x + ab = 0,

x 1 = a, x 2 =b.

Expressing the relationship between the roots and coefficients of equations with general formulas written using symbols, Viète established uniformity in the methods of solving equations. However, the symbolism of Viet is still far from its modern form. He did not recognize negative numbers and therefore, when solving equations, he considered only cases where all the roots were positive.

2. Methods for solving quadratic equations

https://pandia.ru/text/78/082/images/image002_237.gif" height="952">MOU "Sergievsk Secondary School"

Completed by: Sizikov Stanislav

Teacher:

With. Sergievka, 2007

1. Introduction. Quadratic equations in Ancient Babylon……………….3

2. Quadratic equations in Diaphant…………..………………………….4

3. Quadratic equations in India …………………………………………5

4. Quadratic equations of al-Khorezmi……………………………..6

5. Quadratic equations in Europe XIII - XYII…………………………...7

6. About Vieta’s theorem………………………………………………………..9

7. Ten ways to solve quadratic equations…………………..10

8. Conclusion………………………………………………………………20

9. List of references……………………………………………………...21

Introduction

Quadratic equations

Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, and irrational equations. We all know how to solve quadratic equations, starting in the 8th grade. How did the history of solving quadratic equations originate and develop?

Quadratic equations in Ancient Babylon

The need to solve equations not only of the first, but also of the second degree, back in ancient times, was caused by the need to solve problems related to finding the areas of land plots; earthworks of a military nature, as well as with the development of astronomy and mathematics itself. Quadratic equations could be solved around 2000 BC. e. Babylonians. Using modern algebraic notation, we can say that in their cuneiform texts there are, in addition to incomplete ones, such, for example, complete quadratic equations: x2 + x = , : x2 - x = 14https://pandia.ru/text/78/082 /images/image005_150.gif" width="16" height="41 src=">)2 + 12 = x; Bhaskara writes under the guise

x2- 64X = - 768

and, to complete the left side of this equation to square, adds 322 to both sides, then obtaining: x2- 64x + 322 = - 768 + 1024;

(X- 32)2 = 256; X - 32 = ± 16, xt = 16, xg= 48.

Quadratic equations of al-Khorezmi

Al-Khwarizmi's algebraic treatise gives a classification of linear and quadratic equations. The author counts 6 types of equations, expressing them as follows:

1) “Squares are equal to roots,” i.e. ax2 = in.

2) “Squares are equal to numbers,” i.e. ah2= With.

3) “The roots are equal to the number,” i.e. ah = s.

4) “Squares and numbers are equal to roots,” i.e. ah2+ c = in.

5) “Squares and roots are equal to numbers,” i.e. ah2+ in = s.

6) “Roots and numbers are equal to squares,” i.e. input+ c = ax2. For al-Khwarizmi, who avoided the use of negative numbers, the terms of each of these equations are addends and not subtractables. In this case, equations that do not have positive solutions are obviously not taken into account. The author sets out methods for solving these equations. His decision, of course, does not completely coincide with ours. Not to mention that it is purely rhetorical, it should be noted, for example, that when solving an incomplete quadratic equation of the first type, al-Khorezmi, like all mathematicians until the 17th century, does not take into account the zero solution, probably because in specific practical it doesn't matter in tasks. When solving complete quadratic equations, al-Khwarizmi sets out the rules for solving them using particular numerical examples, and then their geometric proofs.

Let's give an example.

Problem 14. “The square and the number 21 are equal to 10 roots. Find the root" (meaning the root of the equation x2+ 21 = 10X).

Al-Khorezmi's treatise is the first book that has come down to us, which systematically sets out the classification of quadratic equations and gives formulas for their solution.

Quadratic equations in EuropeXIII- XVIIcenturies

Formulas for solving quadratic equations following the model of al-Khwarizmi in Europe were first set forth in the “Book of Abacus” (Fibonacci’s “Book of Abacus”, published in Rome in the middle of the last century, contains 459 pages), written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics from both Islamic countries and Ancient Greece, is distinguished by both completeness and clarity of presentation. The author independently developed some new algebraic examples of problem solving and the first V Europe has approached the introduction of negative numbers. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many problems from the Book of Abacus were used in almost all European textbooks of the 16th-17th centuries. and partly XVIII.

General rule for solving quadratic equations reduced to a single canonical form x2+ in = s, for all possible combinations of coefficient signs in, with was formulated in Europe only in 1544. M. Stiefel.

The derivation of the formula for solving a quadratic equation in general form is available from Viète, but Viète recognized only positive roots. Italian mathematicians Tartaglia, Cardaco, Bombelli were among the first in the 16th century. In addition to positive ones, negative roots are also taken into account. Only in the 17th century. Thanks to the works of Girard, Descartes, Newton and other scientists, the method of solving quadratic equations takes on a modern form.

About Vieta's theorem

The theorem expressing the relationship between the coefficients of a quadratic equation and its roots, named after Vieta, was formulated by him for the first time in 1591 as follows: “If IN+ D, multiplied by A minus A2, equals BD, That A equals IN and equal D».

To understand Vieta, we should remember that A, like any
vowel letter, meant the unknown (our X), vowels
IN,D- coefficients for the unknown. In the language of modern algebra, the above Vieta formulation means: if there is

(A+ c)x - x 2 = ab, x2 - (a+ b) x + ab = 0, x1 = a, x2 = b.

Expressing the relationship between the roots and coefficients of equations with general formulas written using symbols, Viète established uniformity in the methods of solving equations. However, the symbolism of Viet is still far from its modern form. He did not recognize negative numbers and therefore, when solving equations, he considered only cases where all the roots were positive

Ten ways to solve quadratic equations

1. Factoring the left side of the equation

Let's solve the equation x2+ 10X- 24 = 0. Let’s factorize the left side of the equation:

x2 + 10x - 24 = x2 + 12x - 2x - 24 =

X(x + x + 12) = (x + 12)(x - 2).

Therefore, the equation can be rewritten as follows:

( X + 12)(x - 2) = 0.

Since the product is zero, at least one of its factors is zero. Therefore, the left side of the equation vanishes when x = 2, and also at X= - 12. This means that the numbers 2 and - 12 are the roots of the equation x2 + 10x - 24 = 0.

2. Method for selecting a complete square

Let's explain this method with an example.

Let's solve the equation x2 + 6x - 7 = 0. Select a complete square on the left side. To do this, we write the expression x2 + 6x in the following form:

x2 + 6x = x2 + 2*x*3.

In the resulting expression, the first term is the square of the number x, and the second is the double product of x by 3. Therefore, to get a complete square, you need to add 32, since

x2 + 2 x 3 + 32 = (x + 3)2.

Let us now transform the left side of the equation

x2 + 6x - 7 = 0,

adding to it and subtracting 32. We have:

x2 + 6x - 7 = x2 + 2 X 3 +– 7 = (X- = (x – Z)2 - 16 .

Thus, this equation can be written as follows:

(x + = 0, i.e. (x + 3)2 = 16.

Hence, X+ 3 = 4 x1 = 1, or x + 3 = - 4, x2 = - 7.

3. Solving quadratic equations using the formula

Let's multiply both sides of the equation

ah2+ input+ c = 0, a ≠ 0, on 4a and sequentially we have:

4a2 x2 + 4abx+ 4ac = 0,

((2ah)2 + 2 axb + b2 ) - b2 + 4ac= 0,

(2ah +b)2 = b2- 4ac,

2ah+ b= ± https://pandia.ru/text/78/082/images/image006_128.gif" width="71" height="27">, x1.2 =

In the case of a positive discriminant, i.e. v2 - 4ac > 0, equation ah2+ in + s= 0 has two different roots.

If the discriminant is zero, i.e. b2 - 4ac = 0, then the equation ah2+ input+ With= 0 has a single root, x = - https://pandia.ru/text/78/082/images/image009_95.gif" width="14" height="62">Its roots satisfy Vieta’s theorem, which when A= 1 has the form

x1 x2 = q,

x1 + x2 = - R.

From this we can draw the following conclusions (based on the coefficients R And q the signs of the roots can be predicted).

a) If a free member q given equation (1)
positive (q> 0), then the equation has two identical
according to the sign of the root and it depends on the second coefficient R
If R> 0, then both roots are negative if R< 0, then both
roots are positive.

For example,

x2- 3X + 2 = 0; x1= 2 and x2 = 1, since q = 2 > 0 u p = - 3 < 0;

x2 + 8x + 7 = 0; x 1 = - 7 and x2 = - 1, since q= 7 > 0 and R = 8 > 0.

b) If a free member q given equation (1)
negative (q < 0), then the equation has two roots of different sign, and the larger root will be positive if R< 0, or negative if p> 0.

For example,

x2 + 4x - 5 = 0; x1 = - 5 and x2 = 1, since q = - 5 < 0 и R= 4 > 0;

x2 - 8x - 9 = 0; x1 = 9 and x2= - 1, since q = - 9 < и R= - 8 < 0.

5. Solving equations using the “throw” method

Consider the quadratic equation ax2 + inx+ c = 0, where a ≠ 0. Multiplying both sides by A, we get the equation a2x2 +abx+ ac= 0.

Let ah = y, where X=; then we come to the equation

y2+ by+ ac = 0,

equivalent to this one. Its roots y1 And y2 we find using Vieta's theorem. Finally we get x1= https://pandia.ru/text/78/082/images/image012_77.gif" width="24" height="43">.

With this method the coefficient A multiplied by the free term, as if “thrown” to it, which is why it is called "transfer" method. This method is used when the roots of the equation can be easily found using Vieta's theorem and, most importantly, when the discriminant is an exact square.

1. Solve the equation 2x2 - 11x + 15 = 0.

Solution. Let’s “throw” coefficient 2 to the free term, and as a result we get the equation

y2 - 11 at+ 30 = 0.

According to Vieta’s theorem y1 = 5, y2 = 6, hence x1 = https://pandia.ru/text/78/082/images/image014_69.gif" width="16 height=41" height="41">, t e.

x1 = 2.5 x2 = 3.

Answer: 2,5; 3.

6. Properties of quadratic coefficientsequations

A. Let a quadratic equation be given

ax2 + inx + c= 0, where A ≠ 0.

1. If a + in + c= 0 (i.e. the sum of the coefficients of the equation is zero), then x1 = 1, x2 = .

2. If a - b + c= 0, orb = A + s, then x1 = - 1, X 2 = - https://pandia.ru/text/78/082/images/image016_58.gif" width="44 height=41" height="41">.

Answer: 1; 184">

The following cases are possible:

A straight line and a parabola can intersect at two points, the abscissas of the intersection points are the roots of the quadratic equation;

A straight line and a parabola can touch (only one common point), i.e. the equation has one solution;

A straight line and a parabola do not have common points, that is, a quadratic equation has no roots.

Examples.

1. Solve graphically the equation x2 - 3x - 4 = 0 (Fig. 2).

Solution. Let's write the equation in the form x2 = 3x + 4.

Let's build a parabola y = x2 and direct y = 3x + 4. Direct at= 3x + 4 can be constructed from two points M(0; 4) and N(3; 13). A straight line and a parabola intersect at two points A to B with abscissas x1= - 1 and x2 = 4.

Answer: x1= - 1, x, = 4.

8. Solving quadratic equations using a compass and ruler

We propose the following method for finding the roots of a quadratic equation

ah2+ input+ With= 0

using a compass and ruler (Fig.).

Let us assume that the desired circle intersects the abscissa axis at the points B(x1; 0) and D(x2 ; 0), where x1 And x2- roots of the equation ax2 + inx+With=0,
and passes through points A(0; 1) and C(0; ) on the ordinate axis..gif" width="197" height="123">

So: 1) construct the points https://pandia.ru/text/78/082/images/image023_40.gif" width="171" height="45"> the circle intersects the OX axis at point B(x1;0), and D(x1 ; 0), where x1 and x2 - roots of the quadratic equation ax2+bx+c = 0.

2) The radius of the circle is equal to the ordinate of the center , the circle touches the Ox axis at point B(x1;0), where xx- the root of a quadratic equation.

3) The radius of the circle is less than the ordinate of the center left">

https://pandia.ru/text/78/082/images/image029_34.gif" width="612" height="372">40" height="14">

https://pandia.ru/text/78/082/images/image031_28.gif" width="612" height="432 src=">

From where after substitutions and

simplifications, the equation z2+pz+q=0 follows, and the letter z means the label of any point on the curvilinear scale.

10. Geometric method for solving quadratic equations

In ancient times, when geometry was more developed than algebra, quadratic equations were solved not algebraically, but geometrically. Let us give a famous example from al-Khwarizmi’s Algebra.

And four attached squares i.e. S=x2+10x+25. Replacing x2+10x with 39, we get that S = 39 + 25 = 64, which means that the side of the square ABCD, i.e. segment AB= 8. For the required side X we get the original square

Conclusion

We all know how to solve quadratic equations, from school to graduation. But in the school mathematics course, formulas for the roots of quadratic equations are studied, with the help of which you can solve any quadratic equations. However, having studied this issue more deeply, I became convinced that there are other ways to solve quadratic equations that allow you to solve many equations very quickly and rationally.

Maybe mathematics is somewhere out there in other dimensions, invisible to the eye - everything is written down and we are just getting new facts from the hole with the worlds? ...God knows; but it turns out that if physicists, chemists, economists or archaeologists need a new model of the structure of the world, this model can always be taken from the shelf where mathematicians put it three hundred years ago, or assembled from parts lying on the same shelf. Perhaps these parts will have to be twisted, adjusted to each other, polished, quickly turned out a couple of new theorem bushings; but the theory of the result will not only describe the actual situation, but also predict the consequences! ...

It's a strange thing - this game of the mind that is always right...

Literature

1. Alimov SHA., Ilyin VA. and others. Algebra, 6-8. Trial textbook for grades 6-8 of secondary school. - M., Education, 1981.

2. Bradys math tables for high school. Ed. 57th. - M., Education, 1990. P. 83.

3. Zlotsky - tasks for teaching mathematics. Book for teachers. - M., Education, 1992.

4.M., Mathematics (supplement to the newspaper “First of September”), No. 21/96, 10/97, 24/97, 18/98, 21/98.

5. Okunev functions, equations and inequalities. Teacher's manual. - M., Education, 1972.

6. Solomnik B. C., Sweet questions and problems in mathematics. Ed. 4th, additional - M., Higher School, 1973.

7.M., Mathematics (supplement to the newspaper “First of September”), No. 40, 2000.

Review

for the work of an 11th grade student of the Sergievskaya Secondary Municipal Educational Institution

comprehensive school"

Slide 1

Slide 2

Course objectives: Introduction to new methods for solving quadratic equations Deepening knowledge on the topic “Quadratic Equations” Development of mathematical, intellectual abilities, research skills Creating conditions for personal self-realization

Slide 3

Course objectives: To introduce students to new ways of solving quadratic equations To strengthen the ability to solve equations using known methods To introduce theorems that allow solving equations in non-standard ways To continue the formation of general educational skills and mathematical culture To promote the formation of interest in research activities To create conditions for students to realize and develop interest in the subject of mathematics Prepare students for the right choice of major

Slide 4

Contents of the program Topic 1. Introduction. 1 hour. Definition of a quadratic equation. Full and incomplete sq. equations Methods for solving them. Questioning. Topic 2. Solving the square. equations. Method of factorization Method of extracting a complete square Solution of the square. equations using formulas Solution sq. equations by transfer method Solution sq. equations using T. Vieta Solving sq. equations using coefficient Solution sq. equations graphically Solving sq. equations using compass and ruler Solving sq. equations using a geometric method Solving sq. equations using “nomograms”

Slide 5

A little history... Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. Quadratic equations in Ancient Babylon. Quadratic equations in India. Quadratic equations in al-Khorezmi. Quadratic equations in Europe XIII - XVII centuries.

Slide 6

Slide 7

Slide 8

Slide 9

Slide 10

The famous French scientist Francois Viète (1540-1603) was a lawyer by profession. He devoted his free time to astronomy. Astronomy classes required knowledge of trigonometry and algebra. Viet took up these sciences and soon came to the conclusion about the need to improve them, which he worked on for a number of years. Thanks to his work, algebra becomes the general science of algebraic equations, based on literal calculus. Therefore, it became possible to express the properties of equations and their roots by general formulas.

Slide 11

While doing the work, I noticed: Methods that I will use: Vieta’s theorem Properties of coefficients “throw” method Factorization of the left side into factors Graphical method The methods are interesting, but they take a lot of time and are not always convenient. Graphic method Using a nomogram Rulers and compasses Isolating a complete square I bow to the scientists who discovered these methods and gave science an impetus for development in the topic “Solving quadratic equations”

Slide 12

Factoring the left side of the equation Let's solve the equation x2 + 10x - 24=0. Let's factorize the left side: x2 + 10x - 24= x2 + 12x -2x - 24= x(x + 12) - 2(x + 12)= (x + 12)(x - 2). (x + 12)(x - 2)=0 x + 12=0 or x - 2=0 x= -12 x= 2 Answer: x1= -12, x2 = 2. Solve the equations: x2 - x=0 x2 + 2x=0 x2 - 81=0 x2 + 4x + 3=0 x2 + 2x - 3=0

Slide 13

Full square extraction method Solve the equation x2 + 6x - 7=0 x2 + 6x - 7=x2 + 2x3 + 32 - 32 - 7=(x-3)2 - 9- 7= (x-3)2 - 16 (x -3)2 -16=0 (x-3)2 =16 x-3=4 or x-3=-4 x=1 x=-7 Answer: x1=1, x2 =-7. Solve the equations: x2 - 8x+15=0 x2 +12x +20=0 x2 + 4x + 3=0 x2 + 2x - 2=0 x2 - 6x + 8=0

Slide 14

Solving quadratic equations using the formula Basic formulas: If b is odd, then D= b2-4ac and x 1,2=, (if D>0) If b- is even, then D1= and x1,2=, (if D >0) Solve the equations: 2x2 - 5x + 2=0 6x2 + 5x +1=0 4x2 - 5x + 2=0 2x2 - 6x + 4=0 x2 - 18x +17=0 =

Slide 15

Solving equations using the transfer method Let us solve the equation ax2 + bx + c = 0. Let's multiply both sides of the equation by a, we get a2 x2 +abx+ac=0. Let ax = y, whence x = y/a. Then U2 + bу + ac = 0. Its roots are y1 and y2. Finally, x1 = y1 /a, x1 = y2 /a. Let's solve the equation 2x2 -11x + 15=0. Let's transfer coefficient 2 to the free term: Y2 -11y+30=0. According to Vieta's theorem, y1 = 5 and y2 = 6. x1 =5/2 and x2 =6/2 x1 =2.5 and x2 =3 Answer: x1=2.5, x2 =3 Solve the equation: 2x2 -9x +9=0 10x2 -11x + 3=0 3x2 + 11x +6=0 6x2 +5x - 6=0 3x2 +1x - 4=0

Slide 16

Solving equations using Vieta's theorem Let's solve the equation x2 +10x-24=0. Since x1 * x2 = -24 x1 + x2 = -10, then 24 = 2 * 12, but -10 = -12 + 2, which means x1 = -12 x2 = 2 Answer: x1 = 2, x2 = -12. Solve the equations: x2 - 7x - 30 =0 x2 +2x - 15=0 x2 - 7x + 6=0 3x2 - 5x + 2=0 5x2 + 4x - 9=0

Slide 17

Properties of coefficients of a quadratic equation If a+b+c=0, then x2 = 1, x2 = c/a If a – b + c=0, then x2 =-1, x2 = -c/a Solve the equation x2 + 6x - 7= 0 Let’s solve the equation 2x2 + 3x +1= 0 1 + 6 – 7 =0, which means x1=1, x2 = -7/1=-7. 2 - 3+1=0, which means x1= - 1, x2 = -1/2 Answer: x1=1, x2 =-7. Answer: x1=-1, x2 =-1/2. Solve the equations: 5x2 - 7x +2 =0 Solve the equations: 5x2 - 7x -12 =0 11x2 +25x - 36=0 11x2 +25x +14=0 345x2 -137x -208=0 3x2 +5x +2=0 3x2 + 5x - 8=0 5x2 + 4x - 1=0 5x2 + 4x - 9=0 x2 + 4x +3=0