I hope that after studying this article you will learn how to find the roots of a complete quadratic equation.

Using the discriminant, only complete quadratic equations are solved; to solve incomplete quadratic equations, other methods are used, which you will find in the article “Solving incomplete quadratic equations.”

What quadratic equations are called complete? This equations of the form ax 2 + b x + c = 0, where coefficients a, b and c are not equal to zero. So, to solve a complete quadratic equation, we need to calculate the discriminant D.

D = b 2 – 4ac.

Depending on the value of the discriminant, we will write down the answer.

If the discriminant is a negative number (D< 0),то корней нет.

If the discriminant is zero, then x = (-b)/2a. When the discriminant is a positive number (D > 0),

then x 1 = (-b - √D)/2a, and x 2 = (-b + √D)/2a.

For example. Solve the equation x 2– 4x + 4= 0.

D = 4 2 – 4 4 = 0

x = (- (-4))/2 = 2

Answer: 2.

Solve Equation 2 x 2 + x + 3 = 0.

D = 1 2 – 4 2 3 = – 23

Answer: no roots.

Solve Equation 2 x 2 + 5x – 7 = 0.

D = 5 2 – 4 2 (–7) = 81

x 1 = (-5 - √81)/(2 2)= (-5 - 9)/4= – 3.5

x 2 = (-5 + √81)/(2 2) = (-5 + 9)/4=1

Answer: – 3.5; 1.

So let’s imagine the solution of complete quadratic equations using the diagram in Figure 1.

Using these formulas you can solve any complete quadratic equation. You just need to be careful to the equation was written as a polynomial standard view

A x 2 + bx + c, otherwise you may make a mistake. For example, in writing the equation x + 3 + 2x 2 = 0, you can mistakenly decide that

a = 1, b = 3 and c = 2. Then

D = 3 2 – 4 1 2 = 1 and then the equation has two roots. And this is not true. (See solution to example 2 above).

Therefore, if the equation is not written as a polynomial of the standard form, first the complete quadratic equation must be written as a polynomial of the standard form (the monomial with the largest exponent should come first, that is A x 2 , then with less – bx and then a free member With.

When solving the reduced quadratic equation and a quadratic equation with an even coefficient in the second term, you can use other formulas. Let's get acquainted with these formulas. If in a complete quadratic equation the second term has an even coefficient (b = 2k), then you can solve the equation using the formulas shown in the diagram in Figure 2.

A complete quadratic equation is called reduced if the coefficient at x 2 is equal to one and the equation takes the form x 2 + px + q = 0. Such an equation can be given for solution, or it can be obtained by dividing all coefficients of the equation by the coefficient A, standing at x 2 .

Figure 3 shows a diagram for solving the reduced square
equations. Let's look at an example of the application of the formulas discussed in this article.

Example. Solve the equation

3x 2 + 6x – 6 = 0.

Let's solve this equation using the formulas shown in the diagram in Figure 1.

D = 6 2 – 4 3 (– 6) = 36 + 72 = 108

√D = √108 = √(36 3) = 6√3

x 1 = (-6 - 6√3)/(2 3) = (6 (-1- √(3)))/6 = –1 – √3

x 2 = (-6 + 6√3)/(2 3) = (6 (-1+ √(3)))/6 = –1 + √3

Answer: –1 – √3; –1 + √3

You can notice that the coefficient of x in this equation even number, that is, b = 6 or b = 2k, whence k = 3. Then let’s try to solve the equation using the formulas given in the diagram of the figure D 1 = 3 2 – 3 · (– 6) = 9 + 18 = 27

√(D 1) = √27 = √(9 3) = 3√3

x 1 = (-3 - 3√3)/3 = (3 (-1 - √(3)))/3 = – 1 – √3

x 2 = (-3 + 3√3)/3 = (3 (-1 + √(3)))/3 = – 1 + √3

Answer: –1 – √3; –1 + √3. Noticing that all the coefficients in this quadratic equation are divisible by 3 and performing the division, we get the reduced quadratic equation x 2 + 2x – 2 = 0 Solve this equation using the formulas for the reduced quadratic
equations figure 3.

D 2 = 2 2 – 4 (– 2) = 4 + 8 = 12

√(D 2) = √12 = √(4 3) = 2√3

x 1 = (-2 - 2√3)/2 = (2 (-1 - √(3)))/2 = – 1 – √3

x 2 = (-2 + 2√3)/2 = (2 (-1+ √(3)))/2 = – 1 + √3

Answer: –1 – √3; –1 + √3.

As you can see, when solving this equation using different formulas, we received the same answer. Therefore, having thoroughly mastered the formulas shown in the diagram in Figure 1, you will always be able to solve any complete quadratic equation.

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The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. The discriminant allows you to solve any quadratic equation using general formula, which looks like this:

The discriminant formula depends on the degree of the polynomial. The above formula is suitable for solving quadratic equations of the following form:

The discriminant has the following properties that you need to know:

* "D" is 0 when the polynomial has multiple roots (equal roots);

* "D" is a symmetric polynomial with respect to the roots of the polynomial and is therefore a polynomial in its coefficients; moreover, the coefficients of this polynomial are integers regardless of the extension in which the roots are taken.

Let's say we are given a quadratic equation of the following form:

1 equation

According to the formula we have:

Since \, the equation has 2 roots. Let's define them:

Where can I solve an equation using a discriminant online solver?

You can solve the equation on our website https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch the video instructions and find out how to solve the equation on our website. And if you have any questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

An incomplete quadratic equation differs from classical (complete) equations in that its factors or free term are equal to zero. The graphs of such functions are parabolas. Depending on their general appearance, they are divided into 3 groups. The principles of solution for all types of equations are the same.

There is nothing complicated in determining the type of an incomplete polynomial. It is best to consider the main differences using visual examples:

  1. If b = 0, then the equation is ax 2 + c = 0.
  2. If c = 0, then the expression ax 2 + bx = 0 should be solved.
  3. If b = 0 and c = 0, then the polynomial turns into an equality like ax 2 = 0.

The latter case is more of a theoretical possibility and never occurs in knowledge testing tasks, since the only correct value of the variable x in the expression is zero. In the future, methods and examples of solving incomplete quadratic equations of 1) and 2) types will be considered.

General algorithm for searching variables and examples with solutions

Regardless of the type of equation, the solution algorithm is reduced to the following steps:

  1. Reduce the expression to a form convenient for finding roots.
  2. Perform calculations.
  3. Write down the answer.

The easiest way to solve incomplete equations is to factor the left side and leave a zero on the right. Thus, the formula for an incomplete quadratic equation for finding roots is reduced to calculating the value of x for each of the factors.

You can only learn how to solve it in practice, so let’s consider specific example finding the roots of an incomplete equation:

As you can see, in this case b = 0. Let’s factorize the left side and get the expression:

4(x – 0.5) ⋅ (x + 0.5) = 0.

Obviously, the product is equal to zero when at least one of the factors is equal to zero. The values ​​of the variable x1 = 0.5 and (or) x2 = -0.5 meet similar requirements.

In order to easily and quickly cope with the task of decomposition quadratic trinomial into factors, remember the following formula:

If there is no free term in the expression, the problem is greatly simplified. It will be enough just to find and bracket common denominator. For clarity, consider an example of how to solve incomplete quadratic equations of the form ax2 + bx = 0.

Let's take the variable x out of brackets and get the following expression:

x ⋅ (x + 3) = 0.

Guided by logic, we come to the conclusion that x1 = 0, and x2 = -3.

Traditional solution method and incomplete quadratic equations

What happens if you apply the discriminant formula and try to find the roots of a polynomial with coefficients equal to zero? Let's take an example from the collection typical tasks for the Unified State Examination in Mathematics 2017, we will solve it using standard formulas and the factorization method.

7x 2 – 3x = 0.

Let's calculate the discriminant value: D = (-3)2 – 4 ⋅ (-7) ⋅ 0 = 9. It turns out that the polynomial has two roots:

Now, let's solve the equation by factoring and compare the results.

X ⋅ (7x + 3) = 0,

2) 7x + 3 = 0,
7x = -3,
x = -.

As you can see, both methods give the same result, but solving the equation using the second method was much easier and faster.

Vieta's theorem

But what to do with Vieta’s favorite theorem? Is it possible to use this method with an incomplete trinomial? Let's try to understand the aspects of casting incomplete equations to the classical form ax2 + bx + c = 0.

In fact, it is possible to apply Vieta's theorem in this case. It is only necessary to bring the expression to general appearance, replacing the missing terms with zero.

For example, with b = 0 and a = 1, in order to eliminate the possibility of confusion, the task should be written in the form: ax2 + 0 + c = 0. Then the ratio of the sum and product of the roots and factors of the polynomial can be expressed as follows:

Theoretical calculations help to get acquainted with the essence of the issue, and always require the development of skills when solving specific problems. Let's turn again to the reference book of standard tasks for the Unified State Exam and find a suitable example:

Let us write the expression in a form convenient for applying Vieta’s theorem:

x 2 + 0 – 16 = 0.

The next step is to create a system of conditions:

Obviously, the roots of the quadratic polynomial will be x 1 = 4 and x 2 = -4.

Now, let's practice bringing the equation to its general form. Let's take the following example: 1/4× x 2 – 1 = 0

In order to apply Vieta's theorem to an expression, it is necessary to get rid of the fraction. Let’s multiply the left and right sides by 4, and look at the result: x2– 4 = 0. The resulting equality is ready to be solved by Vieta’s theorem, but it is much easier and faster to get the answer by simply moving c = 4 to the right side of the equation: x2 = 4.

To summarize, it should be said that the best way Solving incomplete equations by factoring is the simplest and fastest method. If difficulties arise in the process of searching for roots, you can contact traditional method finding roots through a discriminant.

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

Content

See also: Solving quadratic equations online

Basic formulas

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

We further assume that - real numbers.
Let's 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 quadratic trinomial has the form:
.
If the discriminant is equal to 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 you build graph of a function
,
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 x-axis (axis) at two points ().
When , the graph touches the x-axis at one point ().
When , the graph does not cross the x-axis ().

Useful formulas related to quadratic equation

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

Derivation of the formula for the roots of a quadratic equation

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




,
Where
; .

So, we got the formula for a polynomial of the second degree in the form:
.
This shows 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) .


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

From here we obtain the factorization of the quadratic trinomial:

.

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

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

;
;
.

Example 2

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

Let's write the quadratic equation in general form:
.
Comparing with the original equation (2.1), we find the values ​​of the coefficients:
.
We find 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 usually called a multiple. That is, they believe that there are two equal roots:
.

;
.

Example 3

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

Let's write the quadratic equation in general form:
(1) .
Let's rewrite the original equation (3.1):
.
Comparing with (1), we find the values ​​of the coefficients:
.
We find the discriminant:
.
The discriminant is negative, .

Therefore there are no real roots.
;
;
.

You can find complex roots:


.

Then

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

The graph of this function is a parabola. It does not intersect the x-axis (axis). Therefore there are no real roots.
;
;
.

There are no real roots. Complex roots:

See also: The graph of a quadratic function is a parabola. The solutions (roots) of a quadratic equation are the points of intersection of the parabola with the x-axis. If the parabola described, does not intersect with the x-axis, the equation has no real roots. If a parabola intersects the x-axis at one point (the vertex of the parabola), the equation has one real root (the equation is also said to have two coinciding roots). If a parabola intersects the x-axis at two points, the equation has two real roots.

If the coefficient A positive, the branches of the parabola are directed upward; if negative, the branches of the parabola are directed downward. If the coefficient b is positive, then the vertex of the parabola lies in the left half-plane, if negative - in the right half-plane.

Derivation of the formula for solving a quadratic equation

The formula for solving a quadratic equation can be obtained as follows:

a x 2 + b x+ c = 0
a x 2 + b x = - c

Multiply the equation by 4 a

4a 2 x 2 + 4 ab x = -4 ac
4a 2 x 2 + 4 ab x+ b 2 = -4ac + b 2
(2a x+ b) 2 = b 2 -4ac
2a x+ b= ±$\sqrt(b^2-4 a c)$

Finding the roots of a quadratic equation

A quadratic equation with real coefficients can have from 0 to 2 real roots depending on the value of the discriminant D = b 2 − 4ac:

  • for D > 0 there are two roots, and they are calculated by the formula
  • for D = 0 there is one root (two equal or coinciding roots), multiplicity 2: