More in a simple way. To do this, take z out of brackets. You get: z(az + b) = 0. Factors can be written: z=0 and az + b = 0, since both can result in zero. In the notation az + b = 0, we move the second one to the right with a different sign. From here we get z1 = 0 and z2 = -b/a. These are the roots of the original .

If there is incomplete equation of the form az² + c = 0, in this case they are found by simply transferring the free term to right side equations. Also change its sign. You get the record az² \u003d -s. Express z² = -c/a. Take the root and write down two solutions - a positive and a negative value of the square root.

note

If there are fractional coefficients in the equation, multiply the entire equation by the appropriate factor so as to get rid of the fractions.

Knowing how to solve quadratic equations is necessary for both schoolchildren and students, sometimes it can help an adult in everyday life. There are several specific decision methods.

Solving quadratic equations

A quadratic equation of the form a*x^2+b*x+c=0. Coefficient x is the desired variable, a, b, c - numerical coefficients. Remember that the "+" sign can change to the "-" sign.

In order to solve this equation, you must use the Vieta theorem or find the discriminant. The most common way is to find the discriminant, since for some values ​​of a, b, c it is not possible to use the Vieta theorem.

To find the discriminant (D), you must write the formula D=b^2 - 4*a*c. The value of D can be greater than, less than or equal to zero. If D is greater or less than zero, then there will be two roots, if D = 0, then only one root remains, more precisely, we can say that D in this case has two equivalent roots. Substitute the known coefficients a, b, c into the formula and calculate the value.

After you have found the discriminant, to find x, use the formulas: x(1) = (- b+sqrt(D))/2*a; x(2) = (- b-sqrt(D))/2*a where sqrt is the function to take the square root of the given number. After calculating these expressions, you will find the two roots of your equation, after which the equation is considered solved.

If D is less than zero, then it still has roots. At school, this section is practically not studied. University students should be aware that a negative number appears under the root. We get rid of it by separating the imaginary part, that is, -1 under the root is always equal to the imaginary element "i", which is multiplied by the root with the same positive number. For example, if D=sqrt(-20), after the transformation, D=sqrt(20)*i is obtained. After this transformation, the solution of the equation is reduced to the same finding of the roots, as described above.

Vieta's theorem consists in the selection of x(1) and x(2) values. Two identical equations are used: x(1) + x(2)= -b; x(1)*x(2)=s. Moreover, a very important point is the sign in front of the coefficient b, remember that this sign is opposite to the one in the equation. At first glance, it seems that calculating x(1) and x(2) is very simple, but when solving, you will encounter the fact that the numbers will have to be selected exactly.

Elements for solving quadratic equations

According to the rules of mathematics, some can be factored: (a + x (1)) * (b-x (2)) = 0, if you managed to convert using mathematical formulas In a similar way this quadratic equation, then feel free to write down the answer. x(1) and x(2) will be equal to the adjacent coefficients in brackets, but with the opposite sign.

Also, do not forget about incomplete quadratic equations. You may be missing some of the terms, if so, then all its coefficients are simply equal to zero. If x^2 or x is preceded by nothing, then the coefficients a and b are equal to 1.

The transformation of a complete quadratic equation into an incomplete one looks like this (for the case \(b=0\)):

For cases when \(c=0\) or when both coefficients are equal to zero, everything is similar.

Please note that \(a\) is not equal to zero, it cannot be equal to zero, since in this case it turns into:

Solution of incomplete quadratic equations.

First of all, you need to understand that the incomplete quadratic equation is still, therefore, it can be solved in the same way as the usual quadratic (through). To do this, we simply add the missing component of the equation with a zero coefficient.

Example : Find the roots of the equation \(3x^2-27=0\)
Solution :

		We have an incomplete quadratic equation with the coefficient \(b=0\). That is, we can write the equation in the following form:
\(3x^2+0\cdot x-27=0\)		In fact, here is the same equation as at the beginning, but now it can be solved as an ordinary square. First we write down the coefficients.
\(a=3;\) \(b=0;\) \(c=-27;\)		Calculate the discriminant using the formula \(D=b^2-4ac\)
\(D=0^2-4\cdot3\cdot(-27)=\) \(=0+324=324\)		Let's find the roots of the equation using the formulas \(x_(1)=\)\(\frac(-b+\sqrt(D))(2a)\) and \(x_(2)=\)\(\frac(-b-\sqrt(D) )(2a)\)
\(x_(1)=\) \(\frac(-0+\sqrt(324))(2\cdot3)\)\(=\)\(\frac(18)(6)\) \(=3\) \(x_(2)=\) \(\frac(-0-\sqrt(324))(2\cdot3)\)\(=\)\(\frac(-18)(6)\) \(=-3\)		Write down the answer

Answer : \(x_(1)=3\); \(x_(2)=-3\)

Example : Find the roots of the equation \(-x^2+x=0\)
Solution :

		Again, an incomplete quadratic equation, but now the coefficient \(c\) is equal to zero. We write the equation as complete.

Bibliographic description: Gasanov A. R., Kuramshin A. A., Elkov A. A., Shilnenkov N. V., Ulanov D. D., Shmeleva O. V. Methods for solving quadratic equations // Young scientist. - 2016. - No. 6.1. - S. 17-20..02.2019).



Our project is dedicated to the ways of solving quadratic equations. The purpose of the project: to learn how to solve quadratic equations in ways that are not included in the school curriculum. Task: find all possible ways to solve quadratic equations and learn how to use them yourself and introduce classmates to these methods.

What are "quadratic equations"?

Quadratic equation - equation of the form ax2 + bx + c = 0, where a, b, c- some numbers ( a ≠ 0), x- unknown.

The numbers a, b, c are called the coefficients of the quadratic equation.

• a is called the first coefficient;
• b is called the second coefficient;
• c - free member.

And who was the first to "invent" quadratic equations?

Some algebraic techniques for solving linear and quadratic equations were known as early as 4000 years ago in Ancient Babylon. The found ancient Babylonian clay tablets, dated somewhere between 1800 and 1600 BC, are the earliest evidence of the study of quadratic equations. The same tablets contain methods for solving certain types of quadratic equations.

The need to solve equations not only of the first, but also of the second degree in ancient times was caused by the need to solve problems related to finding the areas of land and earthworks of a military nature, as well as the development of astronomy and mathematics itself.

The rule for solving these equations, stated in the Babylonian texts, coincides essentially with the modern one, but it is not known how the Babylonians came to this rule. Almost all the cuneiform texts found so far give only problems with solutions stated in the form of recipes, with no indication of how they were found. In spite of high level development of algebra in Babylon, in cuneiform texts there is no concept of a negative number and general methods for solving quadratic equations.

Babylonian mathematicians from about the 4th century B.C. used the square complement method to solve equations with positive roots. Around 300 B.C. Euclid came up with a more general geometric solution method. The first mathematician who found solutions to an equation with negative roots in the form of an algebraic formula was an Indian scientist. Brahmagupta(India, 7th century AD).

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

ax2 + bx = c, a>0

In this equation, the coefficients can be negative. Brahmagupta's rule essentially coincides with ours.

In India, public competitions in solving difficult problems were common. In one of the old Indian books, the following is said about such competitions: “As the sun outshines the stars with its brilliance, so scientist man eclipse glory in popular assemblies, offering and solving algebraic problems. Tasks were often dressed in poetic form.

In an algebraic treatise Al-Khwarizmi a classification of linear and quadratic equations is given. The author lists 6 types of equations, expressing them as follows:

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

2) “Squares are equal to number”, i.e. ax2 = c.

3) "The roots are equal to the number", i.e. ax2 = c.

4) “Squares and numbers are equal to roots”, i.e. ax2 + c = bx.

5) “Squares and roots are equal to number”, i.e. ax2 + bx = c.

6) “Roots and numbers are equal to squares”, i.e. bx + c == ax2.

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

Forms for solving quadratic equations on the model of Al-Khwarizmi in Europe were first described in the "Book of the Abacus", written in 1202. Italian mathematician Leonard Fibonacci. The author independently developed some new algebraic examples of problem solving and was the first in Europe to approach the introduction of negative numbers.

This book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many tasks from this book were transferred to almost all European textbooks of the 14th-17th centuries. The general rule for solving quadratic equations reduced to a single canonical form x2 + bx = c with all possible combinations of signs and coefficients b, c, was formulated in Europe in 1544. M. Stiefel.

Derivation of the formula for solving a quadratic equation in general view Viet has, but Viet recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli among the first in the 16th century. take into account, in addition to positive, and negative roots. Only in the XVII century. thanks to the work Girard, Descartes, Newton and other scientists, the way of solving quadratic equations takes on a modern form.

Consider several ways to solve quadratic equations.

Standard ways to solve quadratic equations from school curriculum:

1. Factorization of the left side of the equation.
2. Full square selection method.
3. Solution of quadratic equations by formula.
4. Graphic solution quadratic equation.
5. Solution of equations using Vieta's theorem.

Let us dwell in more detail on the solution of reduced and non-reduced quadratic equations using the Vieta theorem.

Recall that to solve the above quadratic equations, it is enough to find two numbers such that the product of which is equal to the free term, and the sum is equal to the second coefficient with the opposite sign.

Example.x 2 -5x+6=0

You need to find numbers whose product is 6 and the sum is 5. These numbers will be 3 and 2.

Answer: x 1 =2, x 2 =3.

But you can use this method for equations with the first coefficient not equal to one.

Example.3x 2 +2x-5=0

We take the first coefficient and multiply it by the free term: x 2 +2x-15=0

The roots of this equation will be numbers whose product is - 15, and the sum is - 2. These numbers are 5 and 3. To find the roots of the original equation, we divide the obtained roots by the first coefficient.

Answer: x 1 =-5/3, x 2 =1

6. Solution of equations by the method of "transfer".

Consider the quadratic equation ax 2 + bx + c = 0, where a≠0.

Multiplying both its parts by a, we get the equation a 2 x 2 + abx + ac = 0.

Let ax = y, whence x = y/a; then we arrive at the equation y 2 + by + ac = 0, which is equivalent to the given one. We find its roots at 1 and at 2 using the Vieta theorem.

Finally we get x 1 = y 1 /a and x 2 = y 2 /a.

With this method, the coefficient a is multiplied by the free term, as if "transferred" to it, therefore it is called the "transfer" method. This method is used when it is easy to find the roots of an equation using Vieta's theorem and, most importantly, when the discriminant is an exact square.

Example.2x 2 - 11x + 15 = 0.

Let's "transfer" the coefficient 2 to the free term and making the replacement we get the equation y 2 - 11y + 30 = 0.

According to Vieta's inverse theorem

y 1 = 5, x 1 = 5/2, x 1 = 2.5; y 2 ​​= 6, x 2 = 6/2, x 2 = 3.

Answer: x 1 =2.5; X 2 = 3.

7. Properties of the coefficients of a quadratic equation.

Let the quadratic equation ax 2 + bx + c \u003d 0, a ≠ 0 be given.

1. If a + b + c \u003d 0 (i.e., the sum of the coefficients of the equation is zero), then x 1 \u003d 1.

2. If a - b + c \u003d 0, or b \u003d a + c, then x 1 \u003d - 1.

Example.345x 2 - 137x - 208 = 0.

Since a + b + c \u003d 0 (345 - 137 - 208 \u003d 0), then x 1 \u003d 1, x 2 \u003d -208/345.

Answer: x 1 =1; X 2 = -208/345 .

Example.132x 2 + 247x + 115 = 0

Because a-b + c \u003d 0 (132 - 247 + 115 \u003d 0), then x 1 \u003d - 1, x 2 \u003d - 115/132

Answer: x 1 = - 1; X 2 =- 115/132

There are other properties of the coefficients of a quadratic equation. but their usage is more complicated.

8. Solving quadratic equations using a nomogram.

Fig 1. Nomogram

This is an old and currently forgotten method for solving quadratic equations, placed on p. 83 of the collection: Bradis V.M. Four-digit mathematical tables. - M., Education, 1990.

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

The curvilinear scale of the nomogram is built according to the formulas (Fig. 1):

Assuming OS = p, ED = q, OE = a(all in cm), from Fig. 1 similarity of triangles SAN and CDF we get the proportion

whence, after substitutions and simplifications, the equation follows z 2 + pz + q = 0, and the letter z means the label of any point on the curved scale.

Rice. 2 Solving a quadratic equation using a nomogram

Examples.

1) For the equation z 2 - 9z + 8 = 0 the nomogram gives the roots z 1 = 8.0 and z 2 = 1.0

Answer: 8.0; 1.0.

2) Solve the equation using the nomogram

2z 2 - 9z + 2 = 0.

Divide the coefficients of this equation by 2, we get the equation z 2 - 4.5z + 1 = 0.

The nomogram gives the roots z 1 = 4 and z 2 = 0.5.

Answer: 4; 0.5.

9. Geometric method for solving quadratic equations.

Example.X 2 + 10x = 39.

In the original, this problem is formulated as follows: "The square and ten roots are equal to 39."

Consider a square with side x, rectangles are built 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, completing four equal squares in the corners, the side of each of them is 2.5, and the area is 6.25

Rice. 3 Graphical way solution of the equation x 2 + 10x = 39

The area S of square ABCD can be represented as the sum of the areas: the original square x 2, four rectangles (4∙2.5x = 10x) and four attached squares (6.25∙4 = 25), i.e. S \u003d x 2 + 10x \u003d 25. Replacing x 2 + 10x with the number 39, we get that S \u003d 39 + 25 \u003d 64, which implies that the side of the square ABCD, i.e. segment AB \u003d 8. For the desired side x of the original square, we get

10. Solution of equations using Bezout's theorem.

Bezout's theorem. The remainder after dividing the polynomial P(x) by the binomial x - α is equal to P(α) (that is, the value of P(x) at x = α).

If the number α is the root of the polynomial P(x), then this polynomial is divisible by x -α without remainder.

Example.x²-4x+3=0

Р(x)= x²-4x+3, α: ±1,±3, α=1, 1-4+3=0. Divide P(x) by (x-1): (x²-4x+3)/(x-1)=x-3

x²-4x+3=(x-1)(x-3), (x-1)(x-3)=0

x-1=0; x=1, or x-3=0, x=3; Answer: x1 =2, x2 =3.

Conclusion: The ability to quickly and rationally solve quadratic equations is simply necessary for solving more complex equations, for example, fractional rational equations, equations higher degrees, biquadratic equations, and in high school trigonometric, exponential and logarithmic equations. Having studied all the methods found for solving quadratic equations, we can advise classmates, in addition to standard methods, to solve by the transfer method (6) and solve equations by the property of coefficients (7), since they are more accessible for understanding.

Literature:

1. Bradis V.M. Four-digit mathematical tables. - M., Education, 1990.
2. Algebra grade 8: textbook for grade 8. general education institutions Makarychev Yu. N., Mindyuk N. G., Neshkov K. I., Suvorova S. B. ed. S. A. Telyakovsky 15th ed., revised. - M.: Enlightenment, 2015
3. https://en.wikipedia.org/wiki/%D0%9A%D0%B2%D0%B0%D0%B4%D1%80%D0%B0%D1%82%D0%BD%D0%BE%D0 %B5_%D1%83%D1%80%D0%B0%D0%B2%D0%BD%D0%B5%D0%BD%D0%B8%D0%B5
4. Glazer G.I. History of mathematics at school. A guide for teachers. / Ed. V.N. Younger. - M.: Enlightenment, 1964.

Formulas for the roots of a quadratic equation. The cases of real, multiple and complex roots are considered. Factorization square trinomial. Geometric interpretation. Examples of determining roots and factorization.

Basic formulas

Consider the quadratic equation:
(1) .
The roots of a quadratic equation(1) are determined by the formulas:
; .
These formulas can be combined like this:
.
When the roots of the quadratic equation are known, then the polynomial of the second degree can be represented as a product of factors (factored):
.

Further, we assume that are real numbers.
Consider discriminant of a quadratic equation:
.
If the discriminant is positive, then the quadratic equation (1) has two different real roots:
; .
Then the factorization of the square trinomial has the form:
.
If the discriminant zero, , then the quadratic equation (1) has two multiple (equal) real roots:
.
Factorization:
.
If the discriminant is negative, then the quadratic equation (1) has two complex conjugate roots:
;
.
Here is the imaginary unit, ;
and are the real and imaginary parts of the roots:
; .
Then

.

Graphic interpretation

If build function graph
,
which is a parabola, then the points of intersection of the graph with the axis will be the roots of the equation
.
When , the graph intersects the abscissa axis (axis) at two points.
When , the graph touches the x-axis at one point.
When , the graph does not cross the x-axis.

Below are examples of such graphs.

Useful Formulas Related to Quadratic Equation

(f.1) ;
(f.2) ;
(f.3) .

Derivation of the formula for the roots of a quadratic equation

We perform transformations and apply formulas (f.1) and (f.3):




,
where
; .

So, we got the formula for the polynomial of the second degree in the form:
.
From this it can be seen that the equation

performed at
and .
That is, and are the roots of the quadratic equation
.

Examples of determining the roots of a quadratic equation

Example 1

(1.1) .

Solution

.
Comparing with our equation (1.1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
Since the discriminant is positive, the equation has two real roots:
;
;
.

From here we obtain the decomposition of the square trinomial into factors:

.

Graph of the function y = 2 x 2 + 7 x + 3 crosses the x-axis at two points.

Let's plot the function
.
The graph of this function is a parabola. It crosses the x-axis (axis) at two points:
and .
These points are the roots of the original equation (1.1).

Answer

;
;
.

Example 2

Find the roots of a quadratic equation:
(2.1) .

Solution

We write the quadratic equation in general form:
.
Comparing with the original equation (2.1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
Since the discriminant is zero, the equation has two multiple (equal) roots:
;
.

Then the factorization of the trinomial has the form:
.

Graph of the function y = x 2 - 4 x + 4 touches the x-axis at one point.

Let's plot the function
.
The graph of this function is a parabola. It touches the x-axis (axis) at one point:
.
This point is the root of the original equation (2.1). Since this root is factored twice:
,
then such a root is called a multiple. That is, they consider that there are two equal roots:
.

Answer

;
.

Example 3

Find the roots of a quadratic equation:
(3.1) .

Solution

We write the quadratic equation in general form:
(1) .
Let us rewrite the original equation (3.1):
.
Comparing with (1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
The discriminant is negative, . Therefore, there are no real roots.

You can find complex roots:
;
;
.

Then


.

The graph of the function does not cross the x-axis. There are no real roots.

Let's plot the function
.
The graph of this function is a parabola. It does not cross the abscissa (axis). Therefore, there are no real roots.

Answer

There are no real roots. Complex roots:
;
;
.

Quadratic equations are studied in grade 8, so there is nothing complicated here. The ability to solve them is essential.

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a , b and c are arbitrary numbers, and a ≠ 0.

Before studying specific solution methods, we note that all quadratic equations can be divided into three classes:

1. Have no roots;
2. They have exactly one root;
3. They have two different roots.

This is an important difference between quadratic and linear equations, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.

Discriminant

Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac .

This formula must be known by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant, you can determine how many roots a quadratic equation has. Namely:

1. If D< 0, корней нет;
2. If D = 0, there is exactly one root;
3. If D > 0, there will be two roots.

Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people think. Take a look at the examples and you will understand everything yourself:

A task. How many roots do quadratic equations have:

1. x 2 - 8x + 12 = 0;
2. 5x2 + 3x + 7 = 0;
3. x 2 − 6x + 9 = 0.

We write the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16

So, the discriminant is positive, so the equation has two different roots. We analyze the second equation in the same way:
a = 5; b = 3; c = 7;
D \u003d 3 2 - 4 5 7 \u003d 9 - 140 \u003d -131.

The discriminant is negative, there are no roots. The last equation remains:
a = 1; b = -6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.

The discriminant is equal to zero - the root will be one.

Note that coefficients have been written out for each equation. Yes, it's long, yes, it's tedious - but you won't mix up the odds and don't make stupid mistakes. Choose for yourself: speed or quality.

By the way, if you “fill your hand”, after a while you will no longer need to write out all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not so much.

The roots of a quadratic equation

Now let's move on to the solution. If the discriminant D > 0, the roots can be found using the formulas:

The basic formula for the roots of a quadratic equation

When D = 0, you can use any of these formulas - you get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.

1. x 2 - 2x - 3 = 0;
2. 15 - 2x - x2 = 0;
3. x2 + 12x + 36 = 0.

First equation:
x 2 - 2x - 3 = 0 ⇒ a = 1; b = −2; c = -3;
D = (−2) 2 − 4 1 (−3) = 16.

D > 0 ⇒ the equation has two roots. Let's find them:

Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 (−1) 15 = 64.

D > 0 ⇒ the equation again has two roots. Let's find them

\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]

Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.

D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:

As you can see from the examples, everything is very simple. If you know the formulas and be able to count, there will be no problems. Most often, errors occur when negative coefficients are substituted into the formula. Here, again, the technique described above will help: look at the formula literally, paint each step - and get rid of mistakes very soon.

Incomplete quadratic equations

It happens that the quadratic equation is somewhat different from what is given in the definition. For example:

1. x2 + 9x = 0;
2. x2 − 16 = 0.

It is easy to see that one of the terms is missing in these equations. Such quadratic equations are even easier to solve than standard ones: they do not even need to calculate the discriminant. So let's introduce a new concept:

The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.

Of course, a very difficult case is possible when both of these coefficients are equal to zero: b \u003d c \u003d 0. In this case, the equation takes the form ax 2 \u003d 0. Obviously, such an equation has a single root: x \u003d 0.

Let's consider other cases. Let b \u003d 0, then we get an incomplete quadratic equation of the form ax 2 + c \u003d 0. Let's slightly transform it:

Because arithmetic Square root exists only from a non-negative number, the last equality makes sense only for (−c /a ) ≥ 0. Conclusion:

1. If an incomplete quadratic equation of the form ax 2 + c = 0 satisfies the inequality (−c / a ) ≥ 0, there will be two roots. The formula is given above;
2. If (−c / a )< 0, корней нет.

As you can see, the discriminant was not required - there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c / a ) ≥ 0. It is enough to express the value of x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If negative, there will be no roots at all.

Now let's deal with equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factorize the polynomial:

Taking the common factor out of the bracket

The product is equal to zero when at least one of the factors is equal to zero. This is where the roots come from. In conclusion, we will analyze several of these equations:

A task. Solve quadratic equations:

1. x2 − 7x = 0;
2. 5x2 + 30 = 0;
3. 4x2 − 9 = 0.

x 2 − 7x = 0 ⇒ x (x − 7) = 0 ⇒ x 1 = 0; x2 = −(−7)/1 = 7.

5x2 + 30 = 0 ⇒ 5x2 = -30 ⇒ x2 = -6. There are no roots, because the square cannot be equal to a negative number.

4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 \u003d -1.5.