This topic may seem difficult at first due to many not so simple formulas. Not only do they themselves quadratic equations have long records, and 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 won't always look like general formula quadratic equation. 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 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, 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. 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.
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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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 decomposition quadratic trinomial into factors has the form:
.
If the discriminant is equal to zero, then the quadratic equation (1) has two multiple (equal) real roots:
.
Factorization:
.
If the discriminant is negative, then the quadratic equation (1) has two complex conjugate roots:
;
.
Here is the imaginary unit, ;
and are the real and imaginary parts of the roots:
;
.
Then
.
Graphic interpretation
If you build graph of a function
,
which is a parabola, then the points of intersection of the graph with the axis will be the roots of the equation
.
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:
;
;
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:
;
;
.
", 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 member.
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 to solve quadratic equations, a special 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.
x 1;2 =
x 1;2 =
x 1;2 =
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 =
x =
| 6 |
| 2 |
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.
For example, for the trinomial \(3x^2+2x-7\), the discriminant will be equal to \(2^2-4\cdot3\cdot(-7)=4+84=88\). And for the trinomial \(x^2-5x+11\), it will be equal to \((-5)^2-4\cdot1\cdot11=25-44=-19\).
The discriminant is denoted by \(D\) and is often used in solving. Also, by the value of the discriminant, you can understand what the graph approximately looks like (see below).
Discriminant and roots of a quadratic equation
The discriminant value shows the number of quadratic equations:
- if \(D\) is positive, the equation will have two roots;
- if \(D\) is equal to zero – there is only one root;
- if \(D\) is negative, there are no roots.
This does not need to be taught, it is not difficult to come to such a conclusion, simply knowing that from the discriminant (that is, \(\sqrt(D)\) is included in the formula for calculating the roots of a quadratic equation: \(x_(1)=\)\( \frac(-b+\sqrt(D))(2a)\) and \(x_(2)=\)\(\frac(-b-\sqrt(D))(2a)\) Let's look at each case more details.
If the discriminant is positive
In this case, the root of it is some positive number, which means \(x_(1)\) and \(x_(2)\) will have different meanings, because in the first formula \(\sqrt(D)\) is added , and in the second it is subtracted. And we have two different roots.
Example
: Find the roots of the equation \(x^2+2x-3=0\)
Solution
:
Answer : \(x_(1)=1\); \(x_(2)=-3\)
If the discriminant is zero
How many roots will there be if the discriminant is zero? Let's reason.
The root formulas look like this: \(x_(1)=\)\(\frac(-b+\sqrt(D))(2a)\) and \(x_(2)=\)\(\frac(-b- \sqrt(D))(2a)\) . And if the discriminant is zero, then its root is also zero. Then it turns out:
\(x_(1)=\)\(\frac(-b+\sqrt(D))(2a)\) \(=\)\(\frac(-b+\sqrt(0))(2a)\) \(=\)\(\frac(-b+0)(2a)\) \(=\)\(\frac(-b)(2a)\)
\(x_(2)=\)\(\frac(-b-\sqrt(D))(2a)\) \(=\)\(\frac(-b-\sqrt(0))(2a) \) \(=\)\(\frac(-b-0)(2a)\) \(=\)\(\frac(-b)(2a)\)
That is, the values of the roots of the equation will be the same, because adding or subtracting zero does not change anything.
Example
: Find the roots of the equation \(x^2-4x+4=0\)
Solution
:
|
\(x^2-4x+4=0\) |
We write out the coefficients: |
|
|
\(a=1;\) \(b=-4;\) \(c=4;\) |
We calculate the discriminant using the formula \(D=b^2-4ac\) |
|
|
\(D=(-4)^2-4\cdot1\cdot4=\) |
Finding the roots of the equation |
|
|
\(x_(1)=\) \(\frac(-(-4)+\sqrt(0))(2\cdot1)\)\(=\)\(\frac(4)(2)\) \(=2\) \(x_(2)=\) \(\frac(-(-4)-\sqrt(0))(2\cdot1)\)\(=\)\(\frac(4)(2)\) \(=2\) |
|
We got two identical roots, so there is no point in writing them separately - we write them as one. |
Answer : \(x=2\)