Quadratic equation where the discriminant is 0. Examples of solving quadratic equations

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 a general formula, which has the following form:

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.

In this article we will look at solving incomplete quadratic equations.

But first, let’s repeat what equations are called quadratic. An equation of the form ax 2 + bx + c = 0, where x is a variable, and the coefficients a, b and c are some numbers, and a ≠ 0, is called square. As we see, the coefficient for x 2 is not equal to zero, and therefore the coefficients for x or the free term can be equal to zero, in which case we get an incomplete quadratic equation.

There are three types of incomplete quadratic equations:

1) If b = 0, c ≠ 0, then ax 2 + c = 0;

2) If b ≠ 0, c = 0, then ax 2 + bx = 0;

3) If b = 0, c = 0, then ax 2 = 0.

• Let's figure out how to solve equations of the form ax 2 + c = 0.

To solve the equation, we move the free term c to the right side of the equation, we get

ax 2 = ‒s. Since a ≠ 0, we divide both sides of the equation by a, then x 2 = ‒c/a.

If ‒с/а > 0, then the equation has two roots

x = ±√(–c/a) .

If ‒c/a< 0, то это уравнение решений не имеет. Более наглядно решение данных уравнений представлено на схеме.

Let's try to understand with examples how to solve such equations.

Example 1. Solve the equation 2x 2 ‒ 32 = 0.

Answer: x 1 = - 4, x 2 = 4.

Example 2. Solve the equation 2x 2 + 8 = 0.

Answer: the equation has no solutions.

• Let's figure out how to solve it equations of the form ax 2 + bx = 0.

To solve the equation ax 2 + bx = 0, let's factorize it, that is, take x out of brackets, we get x(ax + b) = 0. The product is equal to zero if at least one of the factors is equal to zero. Then either x = 0, or ax + b = 0. Solving the equation ax + b = 0, we get ax = - b, whence x = - b/a. An equation of the form ax 2 + bx = 0 always has two roots x 1 = 0 and x 2 = ‒ b/a. See what the solution to equations of this type looks like in the diagram.

Let's consolidate our knowledge with a specific example.

Example 3. Solve the equation 3x 2 ‒ 12x = 0.

x(3x ‒ 12) = 0

x= 0 or 3x – 12 = 0

Answer: x 1 = 0, x 2 = 4.

• Equations of the third type ax 2 = 0 are solved very simply.

If ax 2 = 0, then x 2 = 0. The equation has two equal roots x 1 = 0, x 2 = 0.

For clarity, let's look at the diagram.

Let us make sure when solving Example 4 that equations of this type can be solved very simply.

Example 4. Solve the equation 7x 2 = 0.

Answer: x 1, 2 = 0.

It is not always immediately clear what type of incomplete quadratic equation we have to solve. Consider the following example.

Example 5. Solve the equation

Let's multiply both sides of the equation by a common denominator, that is, by 30

Let's cut it down

5(5x 2 + 9) – 6(4x 2 – 9) = 90.

Let's open the brackets

25x 2 + 45 – 24x 2 + 54 = 90.

Let's give similar

Let's move 99 from the left side of the equation to the right, changing the sign to the opposite

Answer: no roots.

We looked at how incomplete quadratic equations are solved. I hope that now you will not have any difficulties with such tasks. Be careful when determining the type of incomplete quadratic equation, then you will succeed.

If you have questions on this topic, sign up for my lessons, we will solve the problems that arise together.

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Yakupova M.I. 1

Smirnova Yu.V. 1

1 Municipal budgetary educational institution secondary school No. 11

The text of the work is posted without images and formulas.
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History of quadratic equations

Babylon

The need to solve equations not only of the first degree, but also of the second, in ancient times was caused by the need to solve problems related to finding the areas of land plots, with the development of astronomy and mathematics itself. Quadratic equations could be solved around 2000 BC. e. Babylonians. The rules for solving these equations set out in the Babylonian texts are essentially the same as modern ones, but these texts lack the concept of a negative number and general methods for solving quadratic equations.

Ancient Greece

In Ancient Greece, scientists such as Diophantus, Euclid and Heron also worked on solving quadratic equations. Diophantus Diophantus of Alexandria is an ancient Greek mathematician who presumably lived in the 3rd century AD. The main work of Diophantus is “Arithmetic” in 13 books. Euclid. Euclid is an ancient Greek mathematician, the author of the first theoretical treatise on mathematics that has come down to us, Heron. Heron - Greek mathematician and engineer first in Greece in the 1st century AD. gives a purely algebraic way to solve a quadratic equation

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 (VII century), outlined the general rule for solving quadratic equations reduced to a single canonical form: ax2 + bx = c, a> 0. (1) In equation (1) the coefficients can be negative. Brahmagupta's rule is essentially the same as ours. Public competitions in solving difficult problems were common in India. 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 his glory in public 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.

“A flock of frisky monkeys

And twelve along the vines, having eaten to my heart’s content, had fun

They began to jump, hanging

Part eight of them squared

How many monkeys were there?

I was having fun in the clearing

Tell me, in this pack?

Bhaskara's solution indicates that the author knew that the roots of quadratic equations are two-valued. Bhaskar writes the equation corresponding to the problem as x2 - 64x = - 768 and, in order to complete the left side of this equation to a square, adds 322 to both sides, then obtaining: x2 - b4x + 322 = -768 + 1024, (x - 32)2 = 256, x - 32= ±16, x1 = 16, x2 = 48.

Quadratic equations in 17th century Europe

Formulas for solving quadratic equations modeled after Al-Khorezmi 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 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.

Definition of a quadratic equation

An equation of the form ax 2 + bx + c = 0, where a, b, c are numbers, is called quadratic.

Quadratic equation coefficients

Numbers a, b, c are the coefficients of the quadratic equation. a is the first coefficient (before x²), a ≠ 0; b is the second coefficient (before x); c is the free term (without x).

Which of these equations are not quadratic??

1. 4x² + 4x + 1 = 0;2. 5x - 7 = 0;3. - x² - 5x - 1 = 0;4. 2/x² + 3x + 4 = 0;5. ¼ x² - 6x + 1 = 0;6. 2x² = 0;

7. 4x² + 1 = 0;8. x² - 1/x = 0;9. 2x² - x = 0;10. x² -16 = 0;11. 7x² + 5x = 0;12. -8x²= 0;13. 5x³ +6x -8= 0.

Types of quadratic equations

Reduced is a quadratic equation in which the leading coefficient is equal to one. Such an equation can be obtained by dividing the entire expression by the leading coefficient a:

x 2 + px + q =0, p = b/a, q = c/a

A quadratic equation is called complete if all its coefficients are nonzero.

A quadratic equation is called incomplete in which at least one of the coefficients, except the leading one (either the second coefficient or the free term), is equal to zero.

Methods for solving quadratic equations

Method I General formula for calculating roots

To find the roots of a quadratic equation ax 2 + b + c = 0 In general, you should use the algorithm below:

Calculate the value of the discriminant of a quadratic equation: this is the expression for it D= b 2 - 4ac

Derivation of the formula:

Note: It is obvious that the formula for a root of multiplicity 2 is a special case of the general formula, obtained by substituting the equality D=0 into it, and the conclusion about the absence of real roots at D0, and (displaystyle (sqrt (-1))=i) = i.

The presented method is universal, but it is far from the only one. Solving a single equation can be approached in a variety of ways, with preferences usually depending on the solver. In addition, often for this purpose some of the methods turn out to be much more elegant, simple, and less labor-intensive than the standard one.

II method. Roots of a quadratic equation with an even coefficient b III method. Solving incomplete quadratic equations

IV method. Using partial ratios of coefficients

There are special cases of quadratic equations in which the coefficients are in relationships with each other, making them much easier to solve.

Roots of a quadratic equation in which the sum of the leading coefficient and the free term is equal to the second coefficient

If in a quadratic equation ax 2 + bx + c = 0 the sum of the first coefficient and the free term is equal to the second coefficient: a+b=c, then its roots are -1 and the number opposite to the ratio of the free term to the leading coefficient ( -c/a).

Hence, before solving any quadratic equation, you should check the possibility of applying this theorem to it: compare the sum of the leading coefficient and the free term with the second coefficient.

Roots of a quadratic equation whose sum of all coefficients is zero

If in a quadratic equation the sum of all its coefficients is zero, then the roots of such an equation are 1 and the ratio of the free term to the leading coefficient ( c/a).

Hence, before solving an equation using standard methods, you should check the applicability of this theorem to it: add up all the coefficients of this equation and see if this sum is not equal to zero.

V method. Factoring a quadratic trinomial into linear factors

If the trinomial is of the form (displaystyle ax^(2)+bx+c(anot =0))ax 2 + bx + c(a ≠ 0) can somehow be represented as a product of linear factors (displaystyle (kx+m)(lx+n)=0)(kx + m)(lx + n), then we can find the roots of the equation ax 2 + bx + c = 0- they will be -m/k and n/l, indeed, after all (displaystyle (kx+m)(lx+n)=0Longleftrightarrow kx+m=0cup lx+n=0)(kx + m)(lx + n) = 0 kx + mUlx + n, and having solved the indicated linear equations, we obtain the above. Note that the quadratic trinomial does not always decompose into linear factors with real coefficients: this is possible if the corresponding equation has real roots.

Let's consider some special cases

Using the squared sum (difference) formula

If the quadratic trinomial has the form (displaystyle (ax)^(2)+2abx+b^(2))ax 2 + 2abx + b 2 , then by applying the above formula to it, we can factor it into linear factors and, therefore, find roots:

(ax) 2 + 2abx + b 2 = (ax + b) 2

Isolating the full square of the sum (difference)

The above formula is also used using a method called “selecting the full square of the sum (difference).” In relation to the above quadratic equation with the previously introduced notation, this means the following:

Note: If you notice, this formula coincides with that proposed in the section “Roots of the reduced quadratic equation”, which, in turn, can be obtained from the general formula (1) by substituting the equality a=1. This fact is not just a coincidence: using the described method, albeit with some additional reasoning, one can derive a general formula and also prove the properties of the discriminant.

VI method. Using the direct and inverse Vieta theorem

Vieta's direct theorem (see below in the section of the same name) and its inverse theorem allow you to solve the above quadratic equations orally, without resorting to rather cumbersome calculations using formula (1).

According to the converse theorem, every pair of numbers (number) (displaystyle x_(1),x_(2))x 1, x 2, being a solution to the system of equations below, are the roots of the equation

In the general case, that is, for an unreduced quadratic equation ax 2 + bx + c = 0

x 1 + x 2 = -b/a, x 1 * x 2 = c/a

A direct theorem will help you find numbers that satisfy these equations orally. With its help, you can determine the signs of the roots without knowing the roots themselves. To do this, you should follow the rule:

1) if the free term is negative, then the roots have different signs, and the largest in absolute value of the roots has a sign opposite to the sign of the second coefficient of the equation;

2) if the free term is positive, then both roots have the same sign, and this is the sign opposite to the sign of the second coefficient.

VII method. Transfer method

The so-called “transfer” method allows you to reduce the solution of unreduced and irreducible equations to the form of reduced equations with integer coefficients by dividing them by the leading coefficient to the solution of reduced equations with integer coefficients. It is as follows:

Next, the equation is solved orally in the manner described above, then they return to the original variable and find the roots of the equations (displaystyle y_(1)=ax_(1)) y 1 =ax 1 And y 2 =ax 2 .(displaystyle y_(2)=ax_(2))

Geometric meaning

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

If coefficient (displaystyle a) a positive, the branches of the parabola are directed upward and vice versa. If the coefficient (displaystyle b) bpositive (if positive (displaystyle a) a, if negative, vice versa), then the vertex of the parabola lies in the left half-plane and vice versa.

Application of quadratic equations in life

The quadratic equation is widely used. It is used in many calculations, structures, sports, and also around us.

Let us consider and give some examples of the application of the quadratic equation.

Sport. High jumps: during the jumper's run-up, calculations related to the parabola are used to achieve the clearest possible impact on the take-off bar and high flight.

Also, similar calculations are needed in throwing. The flight range of an object depends on the quadratic equation.

Astronomy. The trajectory of the planets can be found using a quadratic equation.

Airplane flight. Airplane takeoff is the main component of flight. Here we take the calculation for low resistance and acceleration of takeoff.

Quadratic equations are also used in various economic disciplines, in programs for processing audio, video, vector and raster graphics.

Conclusion

As a result of the work done, it turned out that quadratic equations attracted scientists back in ancient times, they have already encountered them when solving some problems and tried to solve them. Looking at different ways to solve quadratic equations, I came to the conclusion that not all of them are simple. In my opinion, the best way to solve quadratic equations is to solve them using formulas. The formulas are easy to remember, this method is universal. The hypothesis that equations are widely used in life and mathematics was confirmed. After studying the topic, I learned many interesting facts about quadratic equations, their use, application, types, solutions. And I will be happy to continue studying them. I hope this will help me do well in my exams.

List of used literature

Site materials:

Wikipedia

Open lesson.rf

Handbook of Elementary Mathematics Vygodsky M. Ya.

Just. According to formulas and clear, simple rules. At the first stage

it is necessary to bring the given equation to a standard form, i.e. to the form:

If the equation is already given to you in this form, you do not need to do the first stage. The most important thing is to do it right

determine all the coefficients, A, b And c.

Formula for finding the roots of a quadratic equation.

The expression under the root sign is called discriminant . As you can see, to find X, we

we use only a, b and c. Those. coefficients from quadratic equation. Just carefully put it in

values a, b and c We calculate into this formula. We substitute with their signs!

For example, in the equation:

A =1; b = 3; c = -4.

We substitute the values ​​and write:

The example is almost solved:

This is the answer.

The most common mistakes are confusion with sign values a, b And With. Or rather, with substitution

negative values ​​into the formula for calculating the roots. A detailed recording of the formula comes to the rescue here

with specific numbers. If you have problems with calculations, do it!

Suppose we need to solve the following example:

Here a = -6; b = -5; c = -1

We describe everything in detail, carefully, without missing anything with all the signs and brackets:

Quadratic equations often look slightly different. For example, like this:

Now take note of practical techniques that dramatically reduce the number of errors.

First appointment. Don't be lazy before solving a quadratic equation bring it to standard form.

What does this mean?

Let's say that after all the transformations you get the following equation:

Don't rush to write the root formula! You'll almost certainly get the odds mixed up a, b and c.

Construct the example correctly. First, X squared, then without square, then the free term. Like this:

Get rid of the minus. How? We need to multiply the entire equation by -1. We get:

But now you can safely write down the formula for the roots, calculate the discriminant and finish solving the example.

Decide for yourself. You should now have roots 2 and -1.

Reception second. Check the roots! By Vieta's theorem.

To solve the given quadratic equations, i.e. if the coefficient

x 2 +bx+c=0,

Thenx 1 x 2 =c

x 1 +x 2 =−b

For a complete quadratic equation in which a≠1:

x 2 +bx+c=0,

divide the whole equation by A:

→ →

Where x 1 And x 2 - roots of the equation.

Reception third. If your equation has fractional coefficients, get rid of the fractions! Multiply

equation with a common denominator.

Conclusion. Practical tips:

1. Before solving, we bring the quadratic equation to standard form and build it Right.

2. If there is a negative coefficient in front of the X squared, we eliminate it by multiplying everything

equations by -1.

3. If the coefficients are fractional, we eliminate the fractions by multiplying the entire equation by the corresponding

factor.

4. If x squared is pure, its coefficient is equal to one, the solution can be easily checked by

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.

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):
.

Next we assume that are 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 plot the function
,
which is a parabola, then the points of intersection of the graph with the axis will be the roots of the equation
.
At , 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.

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 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) .

Solution

.
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).

Answer

;
;
.

Example 2

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

Solution

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). Because this root is factored twice:
,
then such a root is usually called a multiple. That is, they believe that there are two equal roots:
.

Answer

;
.

Example 3

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

Solution

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


.

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 intersect the x-axis (axis). Therefore there are no real roots.

Answer

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