A quadratic equation has one root if. Quadratic equations - examples with solutions, features and formulas

21.10.2019

In modern society, the ability to perform operations with equations containing a squared variable can be useful in many areas of activity and is widely used in practice in scientific and technical developments. Evidence of this can be found in the design of sea and river vessels, aircraft and missiles. Using such calculations, the trajectories of movement of a wide variety of bodies, including space objects, are determined. Examples with the solution of quadratic equations are used not only in economic forecasting, in the design and construction of buildings, but also in the most ordinary everyday circumstances. They may be needed on hiking trips, at sporting events, in stores when making purchases and in other very common situations.

Let's break the expression into its component factors

The degree of an equation is determined by the maximum value of the degree of the variable that the expression contains. If it is equal to 2, then such an equation is called quadratic.

If we speak in the language of formulas, then the indicated expressions, no matter how they look, can always be brought to the form when the left side of the expression consists of three terms. Among them: ax 2 (that is, a variable squared with its coefficient), bx (an unknown without a square with its coefficient) and c (a free component, that is, an ordinary number). All this on the right side is equal to 0. In the case when such a polynomial lacks one of its constituent terms, with the exception of ax 2, it is called an incomplete quadratic equation. Examples with the solution of such problems, the values ​​of the variables in which are easy to find, should be considered first.

If the expression looks like it has two terms on the right side, more precisely ax 2 and bx, the easiest way to find x is by putting the variable out of brackets. Now our equation will look like this: x(ax+b). Next, it becomes obvious that either x=0, or the problem comes down to finding a variable from the following expression: ax+b=0. This is dictated by one of the properties of multiplication. The rule states that the product of two factors results in 0 only if one of them is zero.

Example

x=0 or 8x - 3 = 0

As a result, we get two roots of the equation: 0 and 0.375.

Equations of this kind can describe the movement of bodies under the influence of gravity, which began to move from a certain point taken as the origin of coordinates. Here the mathematical notation takes the following form: y = v 0 t + gt 2 /2. By substituting the necessary values, equating the right side to 0 and finding possible unknowns, you can find out the time that passes from the moment the body rises to the moment it falls, as well as many other quantities. But we'll talk about this later.

Factoring an Expression

The rule described above makes it possible to solve these problems in more complex cases. Let's look at examples of solving quadratic equations of this type.

X 2 - 33x + 200 = 0

This quadratic trinomial is complete. First, let's transform the expression and factor it. There are two of them: (x-8) and (x-25) = 0. As a result, we have two roots 8 and 25.

Examples with solving quadratic equations in grade 9 allow this method to find a variable in expressions not only of the second, but even of the third and fourth orders.

For example: 2x 3 + 2x 2 - 18x - 18 = 0. When factoring the right side into factors with a variable, there are three of them, that is, (x+1), (x-3) and (x+3).

As a result, it becomes obvious that this equation has three roots: -3; -1; 3.

Square Root

Another case of an incomplete second-order equation is an expression represented in the language of letters in such a way that the right-hand side is constructed from the components ax 2 and c. Here, to obtain the value of the variable, the free term is transferred to the right side, and after that the square root is extracted from both sides of the equality. It should be noted that in this case there are usually two roots of the equation. The only exceptions can be equalities that do not contain a term with at all, where the variable is equal to zero, as well as variants of expressions when the right side is negative. In the latter case, there are no solutions at all, since the above actions cannot be performed with roots. Examples of solutions to quadratic equations of this type should be considered.

In this case, the roots of the equation will be the numbers -4 and 4.

Calculation of land area

The need for this kind of calculations appeared in ancient times, because the development of mathematics in those distant times was largely determined by the need to determine with the greatest accuracy the areas and perimeters of land plots.

We should also consider examples of solving quadratic equations based on problems of this kind.

So, let's say there is a rectangular plot of land, the length of which is 16 meters greater than the width. You should find the length, width and perimeter of the site if you know that its area is 612 m2.

To get started, let's first create the necessary equation. Let us denote by x the width of the area, then its length will be (x+16). From what has been written it follows that the area is determined by the expression x(x+16), which, according to the conditions of our problem, is 612. This means that x(x+16) = 612.

Solving complete quadratic equations, and this expression is exactly that, cannot be done in the same way. Why? Although the left side still contains two factors, their product does not equal 0 at all, so different methods are used here.

Discriminant

First of all, we will make the necessary transformations, then the appearance of this expression will look like this: x 2 + 16x - 612 = 0. This means that we have received the expression in a form corresponding to the previously specified standard, where a=1, b=16, c= -612.

This could be an example of solving quadratic equations using a discriminant. Here the necessary calculations are made according to the scheme: D = b 2 - 4ac. This auxiliary quantity not only makes it possible to find the required quantities in a second-order equation, it determines the number of possible options. If D>0, there are two of them; for D=0 there is one root. In case D<0, никаких шансов для решения у уравнения вообще не имеется.

About roots and their formula

In our case, the discriminant is equal to: 256 - 4(-612) = 2704. This suggests that our problem has an answer. If you know k, the solution of quadratic equations must be continued using the formula below. It allows you to calculate the roots.

This means that in the presented case: x 1 =18, x 2 =-34. The second option in this dilemma cannot be a solution, because the dimensions of the land plot cannot be measured in negative quantities, which means x (that is, the width of the plot) is 18 m. From here we calculate the length: 18+16=34, and the perimeter 2(34+ 18)=104(m2).

Examples and tasks

We continue our study of quadratic equations. Examples and detailed solutions of several of them will be given below.

1) 15x 2 + 20x + 5 = 12x 2 + 27x + 1

Let’s move everything to the left side of the equality, make a transformation, that is, we’ll get the type of equation that is usually called standard, and equate it to zero.

15x 2 + 20x + 5 - 12x 2 - 27x - 1 = 0

Adding similar ones, we determine the discriminant: D = 49 - 48 = 1. This means our equation will have two roots. Let's calculate them according to the above formula, which means that the first of them will be equal to 4/3, and the second to 1.

2) Now let's solve mysteries of a different kind.

Let's find out if there are any roots here x 2 - 4x + 5 = 1? To obtain a comprehensive answer, let’s reduce the polynomial to the corresponding usual form and calculate the discriminant. In the above example, it is not necessary to solve the quadratic equation, because this is not the essence of the problem at all. In this case, D = 16 - 20 = -4, which means there really are no roots.

Vieta's theorem

It is convenient to solve quadratic equations using the above formulas and the discriminant, when the square root is taken from the value of the latter. But this does not always happen. However, there are many ways to obtain the values ​​of variables in this case. Example: solving quadratic equations using Vieta's theorem. She is named after who lived in the 16th century in France and made a brilliant career thanks to his mathematical talent and connections at court. His portrait can be seen in the article.

The pattern that the famous Frenchman noticed was as follows. He proved that the roots of the equation add up numerically to -p=b/a, and their product corresponds to q=c/a.

Now let's look at specific tasks.

3x 2 + 21x - 54 = 0

For simplicity, let's transform the expression:

x 2 + 7x - 18 = 0

Let's use Vieta's theorem, this will give us the following: the sum of the roots is -7, and their product is -18. From here we get that the roots of the equation are the numbers -9 and 2. After checking, we will make sure that these variable values ​​really fit into the expression.

Parabola graph and equation

The concepts of quadratic function and quadratic equations are closely related. Examples of this have already been given earlier. Now let's look at some mathematical riddles in a little more detail. Any equation of the described type can be represented visually. Such a relationship, drawn as a graph, is called a parabola. Its various types are presented in the figure below.

Any parabola has a vertex, that is, a point from which its branches emerge. If a>0, they go high to infinity, and when a<0, они рисуются вниз. Простейшим примером подобной зависимости является функция y = x 2 . В данном случае в уравнении x 2 =0 неизвестное может принимать только одно значение, то есть х=0, а значит существует только один корень. Это неудивительно, ведь здесь D=0, потому что a=1, b=0, c=0. Выходит формула корней (точнее одного корня) квадратного уравнения запишется так: x = -b/2a.

Visual representations of functions help solve any equations, including quadratic ones. This method is called graphical. And the value of the x variable is the abscissa coordinate at the points where the graph line intersects with 0x. The coordinates of the vertex can be found using the formula just given x 0 = -b/2a. And by substituting the resulting value into the original equation of the function, you can find out y 0, that is, the second coordinate of the vertex of the parabola, which belongs to the ordinate axis.

The intersection of the branches of a parabola with the abscissa axis

There are a lot of examples of solving quadratic equations, but there are also general patterns. Let's look at them. It is clear that the intersection of the graph with the 0x axis for a>0 is possible only if 0 takes negative values. And for a<0 координата у 0 должна быть положительна. Для указанных вариантов D>0. Otherwise D<0. А когда D=0, вершина параболы расположена непосредственно на оси 0х.

From the graph of the parabola you can also determine the roots. The opposite is also true. That is, if it is not easy to obtain a visual representation of a quadratic function, you can equate the right side of the expression to 0 and solve the resulting equation. And knowing the points of intersection with the 0x axis, it is easier to construct a graph.

From the history

Using equations containing a squared variable, in the old days they not only made mathematical calculations and determined the areas of geometric figures. The ancients needed such calculations for grand discoveries in the fields of physics and astronomy, as well as for making astrological forecasts.

As modern scientists suggest, the inhabitants of Babylon were among the first to solve quadratic equations. This happened four centuries before our era. Of course, their calculations were radically different from those currently accepted and turned out to be much more primitive. For example, Mesopotamian mathematicians had no idea about the existence of negative numbers. They were also unfamiliar with other subtleties that any modern schoolchild knows.

Perhaps even earlier than the scientists of Babylon, the sage from India Baudhayama began solving quadratic equations. This happened about eight centuries before the era of Christ. True, the second-order equations, the methods for solving which he gave, were the simplest. Besides him, Chinese mathematicians were also interested in similar questions in the old days. In Europe, quadratic equations began to be solved only at the beginning of the 13th century, but later they were used in their works by such great scientists as Newton, Descartes and many others.

With this math program you can solve quadratic equation.

The program not only gives the answer to the problem, but also displays the solution process in two ways:
- using a discriminant
- using Vieta's theorem (if possible).

Moreover, the answer is displayed as exact, not approximate.
For example, for the equation \(81x^2-16x-1=0\) the answer is displayed in the following form:

$$ x_1 = \frac(8+\sqrt(145))(81), \quad x_2 = \frac(8-\sqrt(145))(81) $$ and not like this: \(x_1 = 0.247; \quad x_2 = -0.05\)

This program can be useful for high school students in general education schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for parents to control the solution of many problems in mathematics and algebra.

Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with detailed solutions.

In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.

If you are not familiar with the rules for entering a quadratic polynomial, we recommend that you familiarize yourself with them.

Rules for entering a quadratic polynomial
Any Latin letter can act as a variable.

For example: \(x, y, z, a, b, c, o, p, q\), etc.
Numbers can be entered as whole or fractional numbers.

Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.
Rules for entering decimal fractions.
In decimal fractions, the fractional part can be separated from the whole part by either a period or a comma.

For example, you can enter decimal fractions like this: 2.5x - 3.5x^2
Rules for entering ordinary fractions.

Only a whole number can act as the numerator, denominator and integer part of a fraction.

The denominator cannot be negative. /
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: &
The whole part is separated from the fraction by the ampersand sign:
Input: 3&1/3 - 5&6/5z +1/7z^2

Result: \(3\frac(1)(3) - 5\frac(6)(5) z + \frac(1)(7)z^2\) When entering an expression you can use parentheses
. In this case, when solving a quadratic equation, the introduced expression is first simplified.


=0
Example: x^2+2x-1

Decide
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A little theory.

Quadratic equation and its roots. Incomplete quadratic equations

Each of the equations
\(-x^2+6x+1.4=0, \quad 8x^2-7x=0, \quad x^2-\frac(4)(9)=0 \)
looks like
\(ax^2+bx+c=0, \)
where x is a variable, a, b and c are numbers.
In the first equation a = -1, b = 6 and c = 1.4, in the second a = 8, b = -7 and c = 0, in the third a = 1, b = 0 and c = 4/9. Such equations are called quadratic equations.

Definition.
Quadratic equation is called an equation of the form ax 2 +bx+c=0, where x is a variable, a, b and c are some numbers, and \(a \neq 0 \).

The numbers a, b and c are the coefficients of the quadratic equation. The number a is called the first coefficient, the number b is the second coefficient, and the number c is the free term.

In each of the equations of the form ax 2 +bx+c=0, where \(a\neq 0\), the largest power of the variable x is a square. Hence the name: quadratic equation.

Note that a quadratic equation is also called an equation of the second degree, since its left side is a polynomial of the second degree.

A quadratic equation in which the coefficient of x 2 is equal to 1 is called given quadratic equation. For example, the quadratic equations given are the equations
\(x^2-11x+30=0, \quad x^2-6x=0, \quad x^2-8=0 \)

If in a quadratic equation ax 2 +bx+c=0 at least one of the coefficients b or c is equal to zero, then such an equation is called incomplete quadratic equation. Thus, the equations -2x 2 +7=0, 3x 2 -10x=0, -4x 2 =0 are incomplete quadratic equations. In the first of them b=0, in the second c=0, in the third b=0 and c=0.

There are three types of incomplete quadratic equations:
1) ax 2 +c=0, where \(c \neq 0 \);
2) ax 2 +bx=0, where \(b \neq 0 \);
3) ax 2 =0.

Let's consider solving equations of each of these types.

To solve an incomplete quadratic equation of the form ax 2 +c=0 for \(c \neq 0 \), move its free term to the right side and divide both sides of the equation by a:
\(x^2 = -\frac(c)(a) \Rightarrow x_(1,2) = \pm \sqrt( -\frac(c)(a)) \)

Since \(c \neq 0 \), then \(-\frac(c)(a) \neq 0 \)

If \(-\frac(c)(a)>0\), then the equation has two roots.

If \(-\frac(c)(a) To solve an incomplete quadratic equation of the form ax 2 +bx=0 with \(b \neq 0 \) factor its left side and obtain the equation
\(x(ax+b)=0 \Rightarrow \left\( \begin(array)(l) x=0 \\ ax+b=0 \end(array) \right. \Rightarrow \left\( \begin (array)(l) x=0 \\ x=-\frac(b)(a) \end(array) \right.

This means that an incomplete quadratic equation of the form ax 2 +bx=0 for \(b \neq 0 \) always has two roots.

An incomplete quadratic equation of the form ax 2 =0 is equivalent to the equation x 2 =0 and therefore has a single root 0.

Formula for the roots of a quadratic equation

Let us now consider how to solve quadratic equations in which both the coefficients of the unknowns and the free term are nonzero.

Let us solve the quadratic equation in general form and as a result we obtain the formula for the roots. This formula can then be used to solve any quadratic equation.

Solve the quadratic equation ax 2 +bx+c=0

Dividing both sides by a, we obtain the equivalent reduced quadratic equation
\(x^2+\frac(b)(a)x +\frac(c)(a)=0 \)

Let's transform this equation by selecting the square of the binomial:
\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2- \left(\frac(b)(2a)\right)^ 2 + \frac(c)(a) = 0 \Rightarrow \)

\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2 = \left(\frac(b)(2a)\right)^ 2 - \frac(c)(a) \Rightarrow \) \(\left(x+\frac(b)(2a)\right)^2 = \frac(b^2)(4a^2) - \frac( c)(a) \Rightarrow \left(x+\frac(b)(2a)\right)^2 = \frac(b^2-4ac)(4a^2) \Rightarrow \) \(x+\frac(b )(2a) = \pm \sqrt( \frac(b^2-4ac)(4a^2) ) \Rightarrow x = -\frac(b)(2a) + \frac( \pm \sqrt(b^2 -4ac) )(2a) \Rightarrow \) \(x = \frac( -b \pm \sqrt(b^2-4ac) )(2a) \)

The radical expression is called discriminant of a quadratic equation ax 2 +bx+c=0 (“discriminant” in Latin - discriminator). It is designated by the letter D, i.e.
\(D = b^2-4ac\)

Now, using the discriminant notation, we rewrite the formula for the roots of the quadratic equation:
\(x_(1,2) = \frac( -b \pm \sqrt(D) )(2a) \), where \(D= b^2-4ac \)

It's obvious that:
1) If D>0, then the quadratic equation has two roots.
2) If D=0, then the quadratic equation has one root \(x=-\frac(b)(2a)\).
3) If D Thus, depending on the value of the discriminant, a quadratic equation can have two roots (for D > 0), one root (for D = 0) or have no roots (for D When solving a quadratic equation using this formula, it is advisable to do the following way:
1) calculate the discriminant and compare it with zero;
2) if the discriminant is positive or equal to zero, then use the root formula; if the discriminant is negative, then write down that there are no roots.

Vieta's theorem

The given quadratic equation ax 2 -7x+10=0 has roots 2 and 5. The sum of the roots is 7, and the product is 10. We see that the sum of the roots is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term. Any reduced quadratic equation that has roots has this property.

The sum of the roots of the above quadratic equation is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term.

Those. Vieta's theorem states that the roots x 1 and x 2 of the reduced quadratic equation x 2 +px+q=0 have the property:
\(\left\( \begin(array)(l) x_1+x_2=-p \\ x_1 \cdot x_2=q \end(array) \right. \)

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 Equations

(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 this 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

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

Answer

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:

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 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.

This topic may seem complicated at first due to the many not-so-simple formulas. Not only do the quadratic equations themselves have long notations, but the roots are also found through the discriminant. In total, three new formulas are obtained. Not very easy to remember. This is possible only after solving such equations frequently. Then all the formulas will be remembered by themselves.

General view of a quadratic equation

Here we propose their explicit recording, when the largest degree is written first, and then in descending order. There are often situations when the terms are inconsistent. Then it is better to rewrite the equation in descending order of the degree of the variable.

Let us introduce some notation. They are presented in the table below.

If we accept these notations, all quadratic equations are reduced to the following notation.

Moreover, the coefficient a ≠ 0. Let this formula be designated number one.

When an equation is given, it is not clear how many roots there will be in the answer. Because one of three options is always possible:

  • the solution will have two roots;
  • the answer will be one number;
  • the equation will have no roots at all.

And until the decision is finalized, it is difficult to understand which option will appear in a particular case.

Types of recordings of quadratic equations

There may be different entries in tasks. They will not always look like the general quadratic equation formula. Sometimes it will be missing some terms. What was written above is the complete equation. If you remove the second or third term in it, you get something else. These records are also called quadratic equations, only incomplete.

Moreover, only terms with coefficients “b” and “c” can disappear. The number "a" cannot be equal to zero under any circumstances. Because in this case the formula turns into a linear equation. The formulas for the incomplete form of equations will be as follows:

So, there are only two types; in addition to complete ones, there are also incomplete quadratic equations. Let the first formula be number two, and the second - three.

Discriminant and dependence of the number of roots on its value

You need to know this number in order to calculate the roots of the equation. It can always be calculated, no matter what the formula of the quadratic equation is. In order to calculate the discriminant, you need to use the equality written below, which will have number four.

After substituting the coefficient values ​​into this formula, you can get numbers with different signs. If the answer is yes, then the answer to the equation will be two different roots. If the number is negative, there will be no roots of the quadratic equation. If it is equal to zero, there will be only one answer.

How to solve a complete quadratic equation?

In fact, consideration of this issue has already begun. Because first you need to find a discriminant. After it is determined that there are roots of the quadratic equation, and their number is known, you need to use formulas for the variables. If there are two roots, then you need to apply the following formula.

Since it contains a “±” sign, there will be two values. The expression under the square root sign is the discriminant. Therefore, the formula can be rewritten differently.

Formula number five. From the same record it is clear that if the discriminant is equal to zero, then both roots will take the same values.

If solving quadratic equations has not yet been worked out, then it is better to write down the values ​​of all coefficients before applying the discriminant and variable formulas. Later this moment will not cause difficulties. But at the very beginning there is confusion.

How to solve an incomplete quadratic equation?

Everything is much simpler here. There is not even a need for additional formulas. And those that have already been written down for the discriminant and the unknown will not be needed.

First, let's look at incomplete equation number two. In this equality, it is necessary to take the unknown quantity out of brackets and solve the linear equation, which will remain in brackets. The answer will have two roots. The first one is necessarily equal to zero, because there is a multiplier consisting of the variable itself. The second one will be obtained by solving a linear equation.

Incomplete equation number three is solved by moving the number from the left side of the equality to the right. Then you need to divide by the coefficient facing the unknown. All that remains is to extract the square root and remember to write it down twice with opposite signs.

Below are some steps that will help you learn how to solve all kinds of equalities that turn into quadratic equations. They will help the student to avoid mistakes due to inattention. These shortcomings can cause poor grades when studying the extensive topic “Quadratic Equations (8th Grade).” Subsequently, these actions will not need to be performed constantly. Because a stable skill will appear.

  • First you need to write the equation in standard form. That is, first the term with the largest degree of the variable, and then - without a degree, and lastly - just a number.
  • If a minus appears before the coefficient “a”, it can complicate the work for a beginner studying quadratic equations. It's better to get rid of it. For this purpose, all equality must be multiplied by “-1”. This means that all terms will change sign to the opposite.
  • It is recommended to get rid of fractions in the same way. Simply multiply the equation by the appropriate factor so that the denominators cancel out.

Examples

It is required to solve the following quadratic equations:

x 2 − 7x = 0;

15 − 2x − x 2 = 0;

x 2 + 8 + 3x = 0;

12x + x 2 + 36 = 0;

(x+1) 2 + x + 1 = (x+1)(x+2).

The first equation: x 2 − 7x = 0. It is incomplete, therefore it is solved as described for formula number two.

After taking it out of brackets, it turns out: x (x - 7) = 0.

The first root takes the value: x 1 = 0. The second will be found from the linear equation: x - 7 = 0. It is easy to see that x 2 = 7.

Second equation: 5x 2 + 30 = 0. Again incomplete. Only it is solved as described for the third formula.

After moving 30 to the right side of the equation: 5x 2 = 30. Now you need to divide by 5. It turns out: x 2 = 6. The answers will be the numbers: x 1 = √6, x 2 = - √6.

The third equation: 15 − 2x − x 2 = 0. Hereinafter, solving quadratic equations will begin by rewriting them in standard form: − x 2 − 2x + 15 = 0. Now it’s time to use the second useful tip and multiply everything by minus one . It turns out x 2 + 2x - 15 = 0. Using the fourth formula, you need to calculate the discriminant: D = 2 2 - 4 * (- 15) = 4 + 60 = 64. It is a positive number. From what is said above, it turns out that the equation has two roots. They need to be calculated using the fifth formula. It turns out that x = (-2 ± √64) / 2 = (-2 ± 8) / 2. Then x 1 = 3, x 2 = - 5.

The fourth equation x 2 + 8 + 3x = 0 is transformed into this: x 2 + 3x + 8 = 0. Its discriminant is equal to this value: -23. Since this number is negative, the answer to this task will be the following entry: “There are no roots.”

The fifth equation 12x + x 2 + 36 = 0 should be rewritten as follows: x 2 + 12x + 36 = 0. After applying the formula for the discriminant, the number zero is obtained. This means that it will have one root, namely: x = -12/ (2 * 1) = -6.

The sixth equation (x+1) 2 + x + 1 = (x+1)(x+2) requires transformations, which consist in the fact that you need to bring similar terms, first opening the brackets. In place of the first there will be the following expression: x 2 + 2x + 1. After the equality, this entry will appear: x 2 + 3x + 2. After similar terms are counted, the equation will take the form: x 2 - x = 0. It has become incomplete . Something similar to this has already been discussed a little higher. The roots of this will be the numbers 0 and 1.