This topic may seem difficult at first due to 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 most high degree 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 becomes 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 meanings. 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.
Let's first consider incomplete equation at 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.
The 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 actions 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 a broad 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 last - 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, the entire 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, so 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. Here and further, 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 advice 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.
Quadratic equation - easy to solve! *Hereinafter referred to as “KU”. Friends, it would seem that there could be nothing simpler in mathematics than solving such an equation. But something told me that many people have problems with him. I decided to see how many on-demand impressions Yandex gives out per month. Here's what happened, look:
What does it mean? This means that about 70,000 people a month are searching for this information, what does this summer have to do with it, and what will happen among school year— there will be twice as many requests. This is not surprising, because those guys and girls who graduated from school a long time ago and are preparing for the Unified State Exam are looking for this information, and schoolchildren also strive to refresh their memory.
Despite the fact that there are a lot of sites that tell you how to solve this equation, I decided to also contribute and publish the material. Firstly, I would like visitors to come to my site based on this request; secondly, in other articles, when the topic of “KU” comes up, I will provide a link to this article; thirdly, I’ll tell you a little more about his solution than is usually stated on other sites. Let's get started! The content of the article:
A quadratic equation is an equation of the form:
where coefficients a,band c are arbitrary numbers, with a≠0.
IN school course the material is given in the following form - the equations are conditionally divided into three classes:
1. They have two roots.
2. *Have only one root.
3. They have no roots. It is worth especially noting here that they do not have real roots
How are roots calculated? Just!
We calculate the discriminant. Underneath this “terrible” word lies a very simple formula:
The root formulas are as follows:
*You need to know these formulas by heart.
You can immediately write down and solve:
Example:
1. If D > 0, then the equation has two roots.
2. If D = 0, then the equation has one root.
3. If D< 0, то уравнение не имеет действительных корней.
Let's look at the equation:
In this regard, when the discriminant is equal to zero, the school course says that one root is obtained, here it is equal to nine. Everything is correct, it is so, but...
This idea is somewhat incorrect. In fact, there are two roots. Yes, yes, don’t be surprised, you get two equal roots, and to be mathematically precise, then the answer should write two roots:
x 1 = 3 x 2 = 3
But this is so - a small digression. At school you can write it down and say that there is one root.
Now the next example:
As we know, the root of a negative number cannot be taken, so the solutions in in this case No.
That's the whole decision process.
Quadratic function.
This shows what the solution looks like geometrically. This is extremely important to understand (in the future, in one of the articles we will analyze in detail the solution to the quadratic inequality).
This is a function of the form:
where x and y are variables
a, b, c – given numbers, with a ≠ 0
The graph is a parabola:
That is, it turns out that by solving a quadratic equation with “y” equal to zero, we find the points of intersection of the parabola with the x axis. There can be two of these points (the discriminant is positive), one (the discriminant is zero) and none (the discriminant is negative). Details about quadratic function You can view article by Inna Feldman.
Let's look at examples:
Example 1: Solve 2x 2 +8 x–192=0
a=2 b=8 c= –192
D=b 2 –4ac = 8 2 –4∙2∙(–192) = 64+1536 = 1600
Answer: x 1 = 8 x 2 = –12
*It was possible to immediately divide the left and right sides of the equation by 2, that is, simplify it. Calculations will be easier.
Example 2: Decide x 2–22 x+121 = 0
a=1 b=–22 c=121
D = b 2 –4ac =(–22) 2 –4∙1∙121 = 484–484 = 0
We found that x 1 = 11 and x 2 = 11
It is permissible to write x = 11 in the answer.
Answer: x = 11
Example 3: Decide x 2 –8x+72 = 0
a=1 b= –8 c=72
D = b 2 –4ac =(–8) 2 –4∙1∙72 = 64–288 = –224
The discriminant is negative, there is no solution in real numbers.
Answer: no solution
The discriminant is negative. There is a solution!
Here we will talk about solving the equation in the case when a negative discriminant is obtained. Do you know anything about complex numbers? I will not go into detail here about why and where they arose and what their specific role and necessity in mathematics is; this is a topic for a large separate article.
The concept of a complex number.
A little theory.
A complex number z is a number of the form
z = a + bi
where a and b are real numbers, i is the so-called imaginary unit.
a+bi – this is a SINGLE NUMBER, not an addition.
The imaginary unit is equal to the root of minus one:
Now consider the equation:
We get two conjugate roots.
Incomplete quadratic equation.
Let's consider special cases, this is when the coefficient “b” or “c” is equal to zero (or both are equal to zero). They can be solved easily without any discriminants.
Case 1. Coefficient b = 0.
The equation becomes:
Let's transform:
Example:
4x 2 –16 = 0 => 4x 2 =16 => x 2 = 4 => x 1 = 2 x 2 = –2
Case 2. Coefficient c = 0.
The equation becomes:
Let's transform and factorize:
*The product is equal to zero when at least one of the factors is equal to zero.
Example:
9x 2 –45x = 0 => 9x (x–5) =0 => x = 0 or x–5 =0
x 1 = 0 x 2 = 5
Case 3. Coefficients b = 0 and c = 0.
Here it is clear that the solution to the equation will always be x = 0.
Useful properties and patterns of coefficients.
There are properties that allow you to solve equations with large coefficients.
Ax 2 + bx+ c=0 equality holds
a + b+ c = 0, That
- if for the coefficients of the equation Ax 2 + bx+ c=0 equality holds
a+ c =b, That
These properties help to decide a certain type equations
Example 1: 5001 x 2 –4995 x – 6=0
The sum of the odds is 5001+( – 4995)+(– 6) = 0, which means
Example 2: 2501 x 2 +2507 x+6=0
Equality holds a+ c =b, Means
Regularities of coefficients.
1. If in the equation ax 2 + bx + c = 0 the coefficient “b” is equal to (a 2 +1), and the coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal
ax 2 + (a 2 +1)∙x+ a= 0 = > x 1 = –a x 2 = –1/a.
Example. Consider the equation 6x 2 + 37x + 6 = 0.
x 1 = –6 x 2 = –1/6.
2. If in the equation ax 2 – bx + c = 0 the coefficient “b” is equal to (a 2 +1), and the coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal
ax 2 – (a 2 +1)∙x+ a= 0 = > x 1 = a x 2 = 1/a.
Example. Consider the equation 15x 2 –226x +15 = 0.
x 1 = 15 x 2 = 1/15.
3. If in Eq. ax 2 + bx – c = 0 coefficient “b” is equal to (a 2 – 1), and coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal
ax 2 + (a 2 –1)∙x – a= 0 = > x 1 = – a x 2 = 1/a.
Example. Consider the equation 17x 2 +288x – 17 = 0.
x 1 = – 17 x 2 = 1/17.
4. If in the equation ax 2 – bx – c = 0 the coefficient “b” is equal to (a 2 – 1), and the coefficient c is numerically equal to the coefficient “a”, then its roots are equal
ax 2 – (a 2 –1)∙x – a= 0 = > x 1 = a x 2 = – 1/a.
Example. Consider the equation 10x 2 – 99x –10 = 0.
x 1 = 10 x 2 = – 1/10
Vieta's theorem.
Vieta's theorem is named after the famous French mathematician Francois Vieta. Using Vieta's theorem, we can express the sum and product of the roots of an arbitrary KU in terms of its coefficients.
45 = 1∙45 45 = 3∙15 45 = 5∙9.
In total, the number 14 gives only 5 and 9. These are the roots. With a certain skill, using the presented theorem, you can solve many quadratic equations orally immediately.
Vieta's theorem, in addition. It is convenient in that after solving a quadratic equation in the usual way (through a discriminant), the resulting roots can be checked. I recommend doing this always.
TRANSPORTATION METHOD
With this method, the coefficient “a” is multiplied by the free term, as if “thrown” to it, which is why it is called "transfer" method. This method is used when you can easily find the roots of the equation using Vieta's theorem and, most importantly, when the discriminant is an exact square.
If A± b+c≠ 0, then the transfer technique is used, for example:
2X 2 – 11x+ 5 = 0 (1) => X 2 – 11x+ 10 = 0 (2)
Using Vieta's theorem in equation (2), it is easy to determine that x 1 = 10 x 2 = 1
The resulting roots of the equation must be divided by 2 (since the two were “thrown” from x 2), we get
x 1 = 5 x 2 = 0.5.
What is the rationale? Look what's happening.
The discriminants of equations (1) and (2) are equal:
If you look at the roots of the equations, you only get different denominators, and the result depends precisely on the coefficient of x 2:
The second (modified) one has roots that are 2 times larger.
Therefore, we divide the result by 2.
*If we reroll the three, we will divide the result by 3, etc.
Answer: x 1 = 5 x 2 = 0.5
Sq. ur-ie and Unified State Examination.
I’ll tell you briefly about its importance - YOU MUST BE ABLE TO DECIDE quickly and without thinking, you need to know the formulas of roots and discriminants by heart. Many of the problems included in the Unified State Examination tasks boil down to solving a quadratic equation (geometric ones included).
Something worth noting!
1. The form of writing an equation can be “implicit”. For example, the following entry is possible:
15+ 9x 2 - 45x = 0 or 15x+42+9x 2 - 45x=0 or 15 -5x+10x 2 = 0.
You need to bring him to standard view(so as not to get confused when deciding).
2. Remember that x is an unknown quantity and it can be denoted by any other letter - t, q, p, h and others.
", that is, equations of the first degree. In this lesson we will look at what is called a quadratic equation and how to solve it.
What is a quadratic equation?
Important!
The degree of an equation is determined by the highest degree to which the unknown stands.
If the maximum power in which the unknown is “2”, then you have a quadratic equation.
Examples of quadratic equations
- 5x 2 − 14x + 17 = 0
- −x 2 + x +
= 01 3 - x 2 + 0.25x = 0
- x 2 − 8 = 0
Important! The general form of a quadratic equation looks like this:
A x 2 + b x + c = 0
“a”, “b” and “c” are given numbers.- “a” is the first or highest coefficient;
- “b” is the second coefficient;
- “c” is a free term.
To find “a”, “b” and “c” you need to compare your equation with the general form of the quadratic equation “ax 2 + bx + c = 0”.
Let's practice determining the coefficients "a", "b" and "c" in quadratic equations.
| The equation | Odds | |||
|---|---|---|---|---|
|
||||
|
||||
| 1 |
| 3 |
- a = −1
- b = 1
- c =
1 3
- a = 1
- b = 0.25
- c = 0
- a = 1
- b = 0
- c = −8
How to Solve Quadratic Equations
Unlike linear equations, a special method is used to solve quadratic equations. formula for finding roots.
Remember!
To solve a quadratic equation you need:
- reduce the quadratic equation to general appearance"ax 2 + bx + c = 0".
- That is, only “0” should remain on the right side;
use formula for roots:
Let's look at an example of how to use the formula to find the roots of a quadratic equation. Let's solve a quadratic equation.
X 2 − 3x − 4 = 0 The equation “x 2 − 3x − 4 = 0” has already been reduced to the general form “ax 2 + bx + c = 0” and does not require additional simplifications. To solve it, we just need to apply.
formula for finding the roots of a quadratic equation
Let us determine the coefficients “a”, “b” and “c” for this equation.
Let us determine the coefficients “a”, “b” and “c” for this equation.
Let us determine the coefficients “a”, “b” and “c” for this equation.
Let us determine the coefficients “a”, “b” and “c” for this equation.
x 1;2 =
It can be used to solve any quadratic equation.
In the formula “x 1;2 = ” the radical expression is often replaced
“b 2 − 4ac” for the letter “D” and is called discriminant. The concept of a discriminant is discussed in more detail in the lesson “What is a discriminant”.
Let's look at another example of a quadratic equation.
x 2 + 9 + x = 7x
In this form, it is quite difficult to determine the coefficients “a”, “b” and “c”. Let's first reduce the equation to the general form “ax 2 + bx + c = 0”.
X 2 + 9 + x = 7x
x 2 + 9 + x − 7x = 0
x 2 + 9 − 6x = 0
x 2 − 6x + 9 = 0
Now you can use the formula for the roots.
X 1;2 =
X 1;2 =
X 1;2 =
x 1;2 =
| 6 |
| 2 |
x =
x = 3
Answer: x = 3
There are times when quadratic equations have no roots. This situation occurs when the formula contains a negative number under the root.
Continuing the topic “Solving Equations,” the material in this article will introduce you to quadratic equations.
Let's look at everything in detail: the essence and notation of a quadratic equation, define the accompanying terms, analyze the scheme for solving incomplete and complete equations, get acquainted with the formula of roots and the discriminant, establish connections between the roots and coefficients, and of course we will give a visual solution to practical examples.
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Quadratic equation, its typesDefinition 1 Quadratic equation is an equation written as, Where x– variable, a , b and c– some numbers, while a is not zero.
Often, quadratic equations are also called equations of the second degree, since in essence a quadratic equation is algebraic equation second degree.
Let's give an example to illustrate the given definition: 9 x 2 + 16 x + 2 = 0 ; 7, 5 x 2 + 3, 1 x + 0, 11 = 0, etc. These are quadratic equations.
Definition 2
Numbers a, b and c are the coefficients of the quadratic equation is an equation written as, while the coefficient a is called the first, or senior, or coefficient at x 2, b - the second coefficient, or coefficient at x, A c called a free member.
For example, in the quadratic equation 6 x 2 − 2 x − 11 = 0 the leading coefficient is 6, the second coefficient is − 2 , and the free term is equal to − 11 . Let us pay attention to the fact that when the coefficients b and/or c are negative, then use short form records like 6 x 2 − 2 x − 11 = 0, but not 6 x 2 + (− 2) x + (− 11) = 0.
Let us also clarify this aspect: if the coefficients a and/or b equal 1 or − 1 , then they may not take an explicit part in writing the quadratic equation, which is explained by the peculiarities of writing the indicated numerical coefficients. For example, in the quadratic equation y 2 − y + 7 = 0 the leading coefficient is 1, and the second coefficient is − 1 .
Reduced and unreduced quadratic equations
Based on the value of the first coefficient, quadratic equations are divided into reduced and unreduced.
Definition 3
Reduced quadratic equation is a quadratic equation where the leading coefficient is 1. For other values of the leading coefficient, the quadratic equation is unreduced.
Let's give examples: quadratic equations x 2 − 4 · x + 3 = 0, x 2 − x − 4 5 = 0 are reduced, in each of which the leading coefficient is 1.
9 x 2 − x − 2 = 0- unreduced quadratic equation, where the first coefficient is different from 1 .
Any unreduced quadratic equation can be converted into a reduced equation by dividing both sides by the first coefficient (equivalent transformation). The transformed equation will have the same roots as the given unreduced equation or will also have no roots at all.
Consideration concrete example will allow us to clearly demonstrate the transition from an unreduced quadratic equation to a reduced one.
Example 1
Given the equation 6 x 2 + 18 x − 7 = 0 . It is necessary to convert the original equation into the reduced form.
Solution
According to the above scheme, we divide both sides of the original equation by the leading coefficient 6. Then we get: (6 x 2 + 18 x − 7) : 3 = 0: 3, and this is the same as: (6 x 2) : 3 + (18 x) : 3 − 7: 3 = 0 and further: (6: 6) x 2 + (18: 6) x − 7: 6 = 0. From here: x 2 + 3 x - 1 1 6 = 0 . Thus, an equation equivalent to the given one is obtained.
Answer: x 2 + 3 x - 1 1 6 = 0 .
Complete and incomplete quadratic equations
Let's turn to the definition of a quadratic equation. In it we specified that a ≠ 0. A similar condition is necessary for the equation is an equation written as was precisely square, since at a = 0 it essentially transforms into a linear equation b x + c = 0.
In the case when the coefficients b And c are equal to zero (which is possible, both individually and jointly), the quadratic equation is called incomplete.
Definition 4
Incomplete quadratic equation- such a quadratic equation a x 2 + b x + c = 0, where at least one of the coefficients b And c(or both) is zero.
Complete quadratic equation– a quadratic equation in which all numerical coefficients are not equal to zero.
Let's discuss why the types of quadratic equations are given exactly these names.
When b = 0, the quadratic equation takes the form a x 2 + 0 x + c = 0, which is the same as a x 2 + c = 0. At c = 0 the quadratic equation is written as a x 2 + b x + 0 = 0, which is equivalent a x 2 + b x = 0. At b = 0 And c = 0 the equation will take the form a x 2 = 0. The equations that we obtained differ from the complete quadratic equation in that their left-hand sides do not contain either a term with the variable x, or a free term, or both. Actually, this fact gave the name to this type of equation – incomplete.
For example, x 2 + 3 x + 4 = 0 and − 7 x 2 − 2 x + 1, 3 = 0 are complete quadratic equations; x 2 = 0, − 5 x 2 = 0; 11 x 2 + 2 = 0, − x 2 − 6 x = 0 – incomplete quadratic equations.
Solving incomplete quadratic equations
The definition given above makes it possible to distinguish the following types of incomplete quadratic equations:
- a x 2 = 0, this equation corresponds to the coefficients b = 0 and c = 0 ;
- a · x 2 + c = 0 at b = 0 ;
- a · x 2 + b · x = 0 at c = 0.
Let us consider sequentially the solution of each type of incomplete quadratic equation.
Solution of the equation a x 2 =0
As mentioned above, this equation corresponds to the coefficients b And c, equal to zero. The equation a x 2 = 0 can be converted into an equivalent equation x 2 = 0, which we get by dividing both sides of the original equation by the number a, not equal to zero. The obvious fact is that the root of the equation x 2 = 0 this is zero because 0 2 = 0 . This equation has no other roots, which can be explained by the properties of the degree: for any number p, not equal to zero, the inequality is true p 2 > 0, from which it follows that when p ≠ 0 equality p 2 = 0 will never be achieved.
Definition 5
Thus, for the incomplete quadratic equation a x 2 = 0 there is a single root x = 0.
Example 2
For example, let’s solve an incomplete quadratic equation − 3 x 2 = 0. It is equivalent to the equation x 2 = 0, its only root is x = 0, then the original equation has a single root - zero.
Briefly, the solution is written as follows:
− 3 x 2 = 0, x 2 = 0, x = 0.
Solving the equation a x 2 + c = 0
Next in line is the solution of incomplete quadratic equations, where b = 0, c ≠ 0, that is, equations of the form a x 2 + c = 0. Let's transform this equation by moving a term from one side of the equation to the other, changing the sign to the opposite one and dividing both sides of the equation by a number that is not equal to zero:
- transfer c to the right hand side, which gives the equation a x 2 = − c;
- divide both sides of the equation by a, we end up with x = - c a .
Our transformations are equivalent; accordingly, the resulting equation is also equivalent to the original one, and this fact makes it possible to draw conclusions about the roots of the equation. From what the values are a And c the value of the expression - c a depends: it can have a minus sign (for example, if a = 1 And c = 2, then - c a = - 2 1 = - 2) or a plus sign (for example, if a = − 2 And c = 6, then - c a = - 6 - 2 = 3); it is not zero because c ≠ 0. Let us dwell in more detail on situations when - c a< 0 и - c a > 0 .
In the case when - c a< 0 , уравнение x 2 = - c a не будет иметь корней. Утверждая это, мы опираемся на то, что квадратом любого числа является число неотрицательное. Из сказанного следует, что при - c a < 0 ни для какого числа p the equality p 2 = - c a cannot be true.
Everything is different when - c a > 0: remember the square root, and it will become obvious that the root of the equation x 2 = - c a will be the number - c a, since - c a 2 = - c a. It is not difficult to understand that the number - - c a is also the root of the equation x 2 = - c a: indeed, - - c a 2 = - c a.
The equation will have no other roots. We can demonstrate this using the method of contradiction. To begin with, let us define the notations for the roots found above as x 1 And − x 1. Let us assume that the equation x 2 = - c a also has a root x 2, which is different from the roots x 1 And − x 1. We know that by substituting into the equation x its roots, we transform the equation into a fair numerical equality.
For x 1 And − x 1 we write: x 1 2 = - c a , and for x 2- x 2 2 = - c a . Based on the properties of numerical equalities, we subtract one correct equality term by term from another, which will give us: x 1 2 − x 2 2 = 0. We use the properties of operations with numbers to rewrite the last equality as (x 1 − x 2) · (x 1 + x 2) = 0. It is known that the product of two numbers is zero if and only if at least one of the numbers is zero. From the above it follows that x 1 − x 2 = 0 and/or x 1 + x 2 = 0, which is the same x 2 = x 1 and/or x 2 = − x 1. An obvious contradiction arose, because at first it was agreed that the root of the equation x 2 differs from x 1 And − x 1. So, we have proven that the equation has no roots other than x = - c a and x = - - c a.
Let us summarize all the arguments above.
Definition 6
Incomplete quadratic equation a x 2 + c = 0 is equivalent to the equation x 2 = - c a, which:
- will have no roots at - c a< 0 ;
- will have two roots x = - c a and x = - - c a for - c a > 0.
Let us give examples of solving the equations a x 2 + c = 0.
Example 3
Given a quadratic equation 9 x 2 + 7 = 0. It is necessary to find a solution.
Solution
Let's move the free term to the right side of the equation, then the equation will take the form 9 x 2 = − 7.
Let us divide both sides of the resulting equation by 9
, we arrive at x 2 = - 7 9 . On the right side we see a number with a minus sign, which means: y given equation no roots. Then the original incomplete quadratic equation 9 x 2 + 7 = 0 will have no roots.
Answer: the equation 9 x 2 + 7 = 0 has no roots.
Example 4
The equation needs to be solved − x 2 + 36 = 0.
Solution
Let's move 36 to the right side: − x 2 = − 36.
Let's divide both parts by − 1
, we get x 2 = 36. On the right side there is a positive number, from which we can conclude that
x = 36 or
x = - 36 .
Let's extract the root and write down the final result: incomplete quadratic equation − x 2 + 36 = 0 has two roots x=6 or x = − 6.
Answer: x=6 or x = − 6.
Solution of the equation a x 2 +b x=0
Let us analyze the third type of incomplete quadratic equations, when c = 0. To find a solution to an incomplete quadratic equation a x 2 + b x = 0, we will use the factorization method. Let's factorize the polynomial that is on the left side of the equation, taking the common factor out of brackets x. This step will make it possible to transform the original incomplete quadratic equation into its equivalent x (a x + b) = 0. And this equation, in turn, is equivalent to a set of equations x = 0 And a x + b = 0. The equation a x + b = 0 linear, and its root: x = − b a.
Definition 7
Thus, the incomplete quadratic equation a x 2 + b x = 0 will have two roots x = 0 And x = − b a.
Let's reinforce the material with an example.
Example 5
It is necessary to find a solution to the equation 2 3 · x 2 - 2 2 7 · x = 0.
Solution
We'll take it out x outside the brackets we get the equation x · 2 3 · x - 2 2 7 = 0 . This equation is equivalent to the equations x = 0 and 2 3 x - 2 2 7 = 0. Now you should solve the resulting linear equation: 2 3 · x = 2 2 7, x = 2 2 7 2 3.
Briefly write the solution to the equation as follows:
2 3 x 2 - 2 2 7 x = 0 x 2 3 x - 2 2 7 = 0
x = 0 or 2 3 x - 2 2 7 = 0
x = 0 or x = 3 3 7
Answer: x = 0, x = 3 3 7.
Discriminant, formula for the roots of a quadratic equation
To find solutions to quadratic equations, there is a root formula:
Definition 8
x = - b ± D 2 · a, where D = b 2 − 4 a c– the so-called discriminant of a quadratic equation.
Writing x = - b ± D 2 · a essentially means that x 1 = - b + D 2 · a, x 2 = - b - D 2 · a.
It would be useful to understand how this formula was derived and how to apply it.
Derivation of the formula for the roots of a quadratic equation
Let us be faced with the task of solving a quadratic equation is an equation written as. Let us carry out a number of equivalent transformations:
- divide both sides of the equation by a number a, different from zero, we obtain the following quadratic equation: x 2 + b a · x + c a = 0 ;
- let's highlight perfect square on the left side of the resulting equation:
x 2 + b a · x + c a = x 2 + 2 · b 2 · a · x + b 2 · a 2 - b 2 · a 2 + c a = = x + b 2 · a 2 - b 2 · a 2 + c a
After this, the equation will take the form: x + b 2 · a 2 - b 2 · a 2 + c a = 0; - Now it is possible to transfer the last two terms to the right side, changing the sign to the opposite, after which we get: x + b 2 · a 2 = b 2 · a 2 - c a ;
- Finally, we transform the expression written on the right side of the last equality:
b 2 · a 2 - c a = b 2 4 · a 2 - c a = b 2 4 · a 2 - 4 · a · c 4 · a 2 = b 2 - 4 · a · c 4 · a 2 .
Thus, we arrive at the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2 , equivalent to the original equation is an equation written as.
We examined the solution of such equations in the previous paragraphs (solving incomplete quadratic equations). The experience already gained makes it possible to draw a conclusion regarding the roots of the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2:
- with b 2 - 4 a c 4 a 2< 0 уравнение не имеет действительных решений;
- when b 2 - 4 · a · c 4 · a 2 = 0 the equation is x + b 2 · a 2 = 0, then x + b 2 · a = 0.
From here the only root x = - b 2 · a is obvious;
- for b 2 - 4 · a · c 4 · a 2 > 0, the following will be true: x + b 2 · a = b 2 - 4 · a · c 4 · a 2 or x = b 2 · a - b 2 - 4 · a · c 4 · a 2 , which is the same as x + - b 2 · a = b 2 - 4 · a · c 4 · a 2 or x = - b 2 · a - b 2 - 4 · a · c 4 · a 2 , i.e. the equation has two roots.
It is possible to conclude that the presence or absence of roots of the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2 (and therefore the original equation) depends on the sign of the expression b 2 - 4 · a · c 4 · a 2 written on the right side. And the sign of this expression is given by the sign of the numerator, (denominator 4 a 2 will always be positive), that is, the sign of the expression b 2 − 4 a c. This expression b 2 − 4 a c the name is given - the discriminant of the quadratic equation and the letter D is defined as its designation. Here you can write down the essence of the discriminant - based on its value and sign, they can conclude whether the quadratic equation will have real roots, and, if so, what is the number of roots - one or two.
Let's return to the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2 . Let's rewrite it using discriminant notation: x + b 2 · a 2 = D 4 · a 2 .
Let us formulate our conclusions again:
Definition 9
- at D< 0 the equation has no real roots;
- at D=0 the equation has a single root x = - b 2 · a ;
- at D > 0 the equation has two roots: x = - b 2 · a + D 4 · a 2 or x = - b 2 · a - D 4 · a 2. Based on the properties of radicals, these roots can be written in the form: x = - b 2 · a + D 2 · a or - b 2 · a - D 2 · a. And, when we expand the modules and reduce the fractions to common denominator, we get: x = - b + D 2 · a, x = - b - D 2 · a.
So, the result of our reasoning was the derivation of the formula for the roots of a quadratic equation:
x = - b + D 2 a, x = - b - D 2 a, discriminant D calculated by the formula D = b 2 − 4 a c.
These formulas make it possible to determine both real roots when the discriminant is greater than zero. When the discriminant is zero, applying both formulas will give the same root as the only solution to the quadratic equation. In the case where the discriminant is negative, if we try to use the formula for the root of a quadratic equation, we will be faced with the need to extract Square root from a negative number, which will take us beyond the real numbers. With a negative discriminant, the quadratic equation will not have real roots, but a pair of complex conjugate roots is possible, determined by the same root formulas we obtained.
Algorithm for solving quadratic equations using root formulas
It is possible to solve a quadratic equation by immediately using the root formula, but this is generally done when it is necessary to find complex roots.
In the majority of cases, it usually means searching not for complex, but for real roots of a quadratic equation. Then it is optimal, before using the formulas for the roots of a quadratic equation, to first determine the discriminant and make sure that it is not negative (otherwise we will conclude that the equation has no real roots), and then proceed to calculate the value of the roots.
The reasoning above makes it possible to formulate an algorithm for solving a quadratic equation.
Definition 10
To solve a quadratic equation is an equation written as, necessary:
- according to the formula D = b 2 − 4 a c find the discriminant value;
- at D< 0 сделать вывод об отсутствии у квадратного уравнения действительных корней;
- for D = 0, find the only root of the equation using the formula x = - b 2 · a ;
- for D > 0, determine two real roots of the quadratic equation using the formula x = - b ± D 2 · a.
Note that when the discriminant is zero, you can use the formula x = - b ± D 2 · a, it will give the same result as the formula x = - b 2 · a.
Let's look at examples.
Examples of solving quadratic equations
Let us give a solution to the examples for different meanings discriminant.
Example 6
We need to find the roots of the equation x 2 + 2 x − 6 = 0.
Solution
Let's write down the numerical coefficients of the quadratic equation: a = 1, b = 2 and c = − 6. Next we proceed according to the algorithm, i.e. Let's start calculating the discriminant, for which we will substitute the coefficients a, b And c into the discriminant formula: D = b 2 − 4 · a · c = 2 2 − 4 · 1 · (− 6) = 4 + 24 = 28 .
So we get D > 0, which means that the original equation will have two real roots.
To find them, we use the root formula x = - b ± D 2 · a and, substituting the corresponding values, we get: x = - 2 ± 28 2 · 1. Let us simplify the resulting expression by taking the factor out of the root sign and then reducing the fraction:
x = - 2 ± 2 7 2
x = - 2 + 2 7 2 or x = - 2 - 2 7 2
x = - 1 + 7 or x = - 1 - 7
Answer: x = - 1 + 7 , x = - 1 - 7 .
Example 7
Need to solve a quadratic equation − 4 x 2 + 28 x − 49 = 0.
Solution
Let's define the discriminant: D = 28 2 − 4 · (− 4) · (− 49) = 784 − 784 = 0. With this value of the discriminant, the original equation will have only one root, determined by the formula x = - b 2 · a.
x = - 28 2 (- 4) x = 3.5
Answer: x = 3.5.
Example 8
The equation needs to be solved 5 y 2 + 6 y + 2 = 0
Solution
The numerical coefficients of this equation will be: a = 5, b = 6 and c = 2. We use these values to find the discriminant: D = b 2 − 4 · a · c = 6 2 − 4 · 5 · 2 = 36 − 40 = − 4 . The calculated discriminant is negative, so the original quadratic equation has no real roots.
In the case when the task is to indicate complex roots, we apply the root formula, performing actions with complex numbers:
x = - 6 ± - 4 2 5,
x = - 6 + 2 i 10 or x = - 6 - 2 i 10,
x = - 3 5 + 1 5 · i or x = - 3 5 - 1 5 · i.
Answer: there are no real roots; the complex roots are as follows: - 3 5 + 1 5 · i, - 3 5 - 1 5 · i.
IN school curriculum There is no standard requirement to look for complex roots, therefore, if during the solution the discriminant is determined to be negative, the answer is immediately written down that there are no real roots.
Root formula for even second coefficients
The root formula x = - b ± D 2 · a (D = b 2 − 4 · a · c) makes it possible to obtain another formula, more compact, allowing one to find solutions to quadratic equations with an even coefficient for x (or with a coefficient of the form 2 · n, for example, 2 3 or 14 ln 5 = 2 7 ln 5). Let us show how this formula is derived.
Let us be faced with the task of finding a solution to the quadratic equation a · x 2 + 2 · n · x + c = 0 . We proceed according to the algorithm: we determine the discriminant D = (2 n) 2 − 4 a c = 4 n 2 − 4 a c = 4 (n 2 − a c), and then use the root formula:
x = - 2 n ± D 2 a, x = - 2 n ± 4 n 2 - a c 2 a, x = - 2 n ± 2 n 2 - a c 2 a, x = - n ± n 2 - a · c a .
Let the expression n 2 − a · c be denoted as D 1 (sometimes it is denoted D "). Then the formula for the roots of the quadratic equation under consideration with the second coefficient 2 · n will take the form:
x = - n ± D 1 a, where D 1 = n 2 − a · c.
It is easy to see that D = 4 · D 1, or D 1 = D 4. In other words, D 1 is a quarter of the discriminant. Obviously, the sign of D 1 is the same as the sign of D, which means the sign of D 1 can also serve as an indicator of the presence or absence of roots of a quadratic equation.
Definition 11
Thus, to find a solution to a quadratic equation with a second coefficient of 2 n, it is necessary:
- find D 1 = n 2 − a · c ;
- at D 1< 0 сделать вывод, что действительных корней нет;
- when D 1 = 0, determine the only root of the equation using the formula x = - n a;
- for D 1 > 0, determine two real roots using the formula x = - n ± D 1 a.
Example 9
It is necessary to solve the quadratic equation 5 x 2 − 6 x − 32 = 0.
Solution
We can represent the second coefficient of the given equation as 2 · (− 3) . Then we rewrite the given quadratic equation as 5 x 2 + 2 (− 3) x − 32 = 0, where a = 5, n = − 3 and c = − 32.
Let's calculate the fourth part of the discriminant: D 1 = n 2 − a · c = (− 3) 2 − 5 · (− 32) = 9 + 160 = 169. The resulting value is positive, which means that the equation has two real roots. Let us determine them using the corresponding root formula:
x = - n ± D 1 a, x = - - 3 ± 169 5, x = 3 ± 13 5,
x = 3 + 13 5 or x = 3 - 13 5
x = 3 1 5 or x = - 2
It would be possible to carry out calculations using the usual formula for the roots of a quadratic equation, but in this case the solution would be more cumbersome.
Answer: x = 3 1 5 or x = - 2 .
Simplifying the form of quadratic equations
Sometimes it is possible to optimize the form of the original equation, which will simplify the process of calculating the roots.
For example, the quadratic equation 12 x 2 − 4 x − 7 = 0 is clearly more convenient to solve than 1200 x 2 − 400 x − 700 = 0.
More often, simplification of the form of a quadratic equation is carried out by multiplying or dividing its both sides by a certain number. For example, above we showed a simplified representation of the equation 1200 x 2 − 400 x − 700 = 0, obtained by dividing both sides by 100.
Such a transformation is possible when the coefficients of the quadratic equation are not mutually prime numbers. Then we usually divide both sides of the equation by the largest common divisor absolute values of its coefficients.
As an example, we use the quadratic equation 12 x 2 − 42 x + 48 = 0. Let us determine the GCD of the absolute values of its coefficients: GCD (12, 42, 48) = GCD(GCD (12, 42), 48) = GCD (6, 48) = 6. Let us divide both sides of the original quadratic equation by 6 and obtain the equivalent quadratic equation 2 x 2 − 7 x + 8 = 0.
By multiplying both sides of a quadratic equation, you usually get rid of fractional coefficients. In this case, they multiply by the least common multiple of the denominators of its coefficients. For example, if each part of the quadratic equation 1 6 x 2 + 2 3 x - 3 = 0 is multiplied with LCM (6, 3, 1) = 6, then it will become written in more in simple form x 2 + 4 x − 18 = 0 .
Finally, we note that we almost always get rid of the minus at the first coefficient of a quadratic equation by changing the signs of each term of the equation, which is achieved by multiplying (or dividing) both sides by − 1. For example, from the quadratic equation − 2 x 2 − 3 x + 7 = 0, you can go to its simplified version 2 x 2 + 3 x − 7 = 0.
Relationship between roots and coefficients
The formula for the roots of quadratic equations, already known to us, x = - b ± D 2 · a, expresses the roots of the equation through its numerical coefficients. Based on this formula, we have the opportunity to specify other dependencies between the roots and coefficients.
The most famous and applicable formulas are Vieta’s theorem:
x 1 + x 2 = - b a and x 2 = c a.
In particular, for the given quadratic equation, the sum of the roots is the second coefficient with the opposite sign, and the product of the roots is equal to the free term. For example, by looking at the form of the quadratic equation 3 x 2 − 7 x + 22 = 0, it is possible to immediately determine that the sum of its roots is 7 3 and the product of the roots is 22 3.
You can also find a number of other connections between the roots and coefficients of a quadratic equation. For example, the sum of the squares of the roots of a quadratic equation can be expressed in terms of coefficients:
x 1 2 + x 2 2 = (x 1 + x 2) 2 - 2 x 1 x 2 = - b a 2 - 2 c a = b 2 a 2 - 2 c a = b 2 - 2 a c a 2.
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Quadratic equation problems are studied both in the school curriculum and in universities. They mean equations of the form a*x^2 + b*x + c = 0, where x- variable, a, b, c – constants; a<>0 . The task is to find the roots of the equation.
Geometric meaning of quadratic equation
The graph of a function that is represented by a quadratic equation is a parabola. The solutions (roots) of a quadratic equation are the points of intersection of the parabola with the abscissa (x) axis. It follows that there are three possible cases:
1) the parabola has no points of intersection with the abscissa axis. This means that it is in the upper plane with branches up or the bottom with branches down. In such cases, the quadratic equation has no real roots (it has two complex roots).
2) the parabola has one point of intersection with the Ox axis. Such a point is called the vertex of the parabola, and the quadratic equation at it acquires its minimum or maximum value. In this case, the quadratic equation has one real root (or two identical roots).
3) Last case in practice it is more interesting - there are two points of intersection of the parabola with the abscissa axis. This means that there are two real roots of the equation.
Based on the analysis of the coefficients of the powers of the variables, interesting conclusions can be drawn about the placement of the parabola.
1) If the coefficient a is greater than zero, then the parabola’s branches are directed upward; if it is negative, the parabola’s branches are directed downward.
2) If the coefficient b is greater than zero, then the vertex of the parabola lies in the left half-plane if it takes negative meaning- then on the right.
Derivation of the formula for solving a quadratic equation
Let's transfer the constant from the quadratic equation 
for the equal sign, we get the expression
Multiply both sides by 4a 
To get a complete square on the left, add b^2 on both sides and carry out the transformation
From here we find
Formula for the discriminant and roots of a quadratic equation
The discriminant is the value of the radical expression. If it is positive, then the equation has two real roots, calculated by the formula 
When the discriminant is zero, the quadratic equation has one solution (two coinciding roots), which can be easily obtained from the above formula for D=0. When the discriminant is negative, the equation has no real roots. However, solutions to the quadratic equation are found in the complex plane, and their value is calculated using the formula 
Vieta's theorem
Let's consider two roots of a quadratic equation and construct a quadratic equation on their basis. Vieta's theorem itself easily follows from the notation: if we have a quadratic equation of the form 
then the sum of its roots is equal to the coefficient p taken with the opposite sign, and the product of the roots of the equation is equal to the free term q. The formulaic representation of the above will look like If in a classical equation the constant a is nonzero, then you need to divide the entire equation by it, and then apply Vieta’s theorem.
Factoring quadratic equation schedule
Let the task be set: factor a quadratic equation. To do this, we first solve the equation (find the roots). Next, we substitute the found roots into the expansion formula for the quadratic equation. This will solve the problem.
Quadratic equation problems
Task 1. Find the roots of a quadratic equation
x^2-26x+120=0 .
Solution: Write down the coefficients and substitute them into the discriminant formula
Root of given value is equal to 14, it is easy to find with a calculator, or remember with frequent use, however, for convenience, at the end of the article I will give you a list of squares of numbers that can often be encountered in such problems.
We substitute the found value into the root formula 
and we get 
Task 2. Solve the equation
2x 2 +x-3=0.
Solution: We have a complete quadratic equation, write out the coefficients and find the discriminant 
Using known formulas we find the roots of the quadratic equation 
Task 3. Solve the equation
9x 2 -12x+4=0.
Solution: We have a complete quadratic equation. Determining the discriminant
We got a case where the roots coincide. Find the values of the roots using the formula 
Task 4. Solve the equation
x^2+x-6=0 .
Solution: In cases where there are small coefficients for x, it is advisable to apply Vieta’s theorem. By its condition we obtain two equations
From the second condition we find that the product must be equal to -6. This means that one of the roots is negative. We have the following possible pair of solutions (-3;2), (3;-2) . Taking into account the first condition, we reject the second pair of solutions.
The roots of the equation are equal
Problem 5. Find the lengths of the sides of a rectangle if its perimeter is 18 cm and its area is 77 cm 2.
Solution: Half the perimeter of a rectangle is equal to the sum of its adjacent sides. Let's denote x as the larger side, then 18-x is its smaller side. The area of the rectangle is equal to the product of these lengths:
x(18-x)=77;
or
x 2 -18x+77=0.
Let's find the discriminant of the equation
Calculating the roots of the equation 
If x=11, That 18's=7 , the opposite is also true (if x=7, then 21's=9).
Problem 6. Factor the quadratic equation 10x 2 -11x+3=0.
Solution: Let's calculate the roots of the equation, to do this we find the discriminant 
We substitute the found value into the root formula and calculate 
We apply the formula for decomposing a quadratic equation by roots 
Opening the brackets we obtain an identity.
Quadratic equation with parameter
Example 1. At what parameter values A , does the equation (a-3)x 2 + (3-a)x-1/4=0 have one root?
Solution: By direct substitution of the value a=3 we see that it has no solution. Next, we will use the fact that with a zero discriminant the equation has one root of multiplicity 2. Let's write out the discriminant 
Let's simplify it and equate it to zero
We have obtained a quadratic equation with respect to the parameter a, the solution of which can be easily obtained using Vieta’s theorem. The sum of the roots is 7, and their product is 12. By simple search we establish that the numbers 3,4 will be the roots of the equation. Since we already rejected the solution a=3 at the beginning of the calculations, the only correct one will be - a=4. Thus, when a=4 the equation has one root.
Example 2. At what parameter values A , the equation a(a+3)x^2+(2a+6)x-3a-9=0 has more than one root?
Solution: Let's first consider the singular points, they will be the values a=0 and a=-3. When a=0, the equation will be simplified to the form 6x-9=0; x=3/2 and there will be one root. For a= -3 we obtain the identity 0=0.
Let's calculate the discriminant 
and find the value of a at which it is positive 
From the first condition we get a>3. For the second, we find the discriminant and roots of the equation 
Let us determine the intervals where the function takes positive values. By substituting the point a=0 we get 3>0
.
So, outside the interval (-3;1/3) the function is negative. Don't forget the point a=0, which should be excluded because the original equation has one root in it.
As a result, we obtain two intervals that satisfy the conditions of the problem 
There will be many similar tasks in practice, try to figure out the tasks yourself and do not forget to take into account the conditions that are mutually exclusive. Study well the formulas for solving quadratic equations; they are often needed in calculations in various problems and sciences.