20 quadratic equations. Quadratic equation

Let's consider the problem. The base of the rectangle is 10 cm greater than its height, and its area is 24 cm². Find the height of the rectangle. Let X centimeters is the height of the rectangle, then its base is equal to ( X+10) cm. The area of ​​this rectangle is X(X+ 10) cm². According to the conditions of the problem X(X+ 10) = 24. Opening the brackets and moving the number 24 with the opposite sign into left side equations, we get: X² + 10 X-24 = 0. When solving this problem, an equation was obtained that is called quadratic.

A quadratic equation is an equation of the form

ax ²+ bx+c= 0

Where a, b, c- given numbers, and A≠ 0, and X- unknown.

Odds a, b, c The quadratic equation is usually called: a— the first or highest coefficient, b- second coefficient, c- a free member. For example, in our problem, the leading coefficient is 1, the second coefficient is 10, and the free term is -24. Solving many problems in mathematics and physics comes down to solving quadratic equations.

Solving Quadratic Equations

Complete quadratic equations. The first step is to bring the given equation to standard form ax²+ bx+ c = 0. Let's return to our problem, in which the equation can be written as X(X+ 10) = 24 let’s bring it to standard form, open the brackets X² + 10 X- 24 = 0, we solve this equation using the formula for the roots of a general quadratic equation.

The expression under the root sign in this formula is called the discriminant D = b² - 4 ac

If D>0, then the quadratic equation has two different roots, which can be found using the formula for the roots of a quadratic equation.

If D=0, then the quadratic equation has one root.

If D<0, то квадратное уравнение не имеет действительных корней, т. е. не имеет решения.

Let's substitute the values ​​into our formula A= 1, b= 10, c= -24.

we get D>0, therefore we get two roots.

Let's consider an example where D=0, under this condition there should be one root.

25x² — 30 x+ 9 = 0

Consider an example where D<0, при этом условии решения не должно быть.

2x² + 3 x+ 4 = 0

The number under the root sign (discriminant) is negative; we write the answer as follows: the equation has no real roots.

Solving incomplete quadratic equations

Quadratic equation ax² + bx+ c= 0 is called incomplete if at least one of the coefficients b or c equal to zero. An incomplete quadratic equation is an equation of one of the following types:

ax² = 0,

ax² + c= 0, c≠ 0,

ax² + bx= 0, b≠ 0.

Let's look at a few examples and solve the equation

Dividing both sides of the equation by 5 gives the equation X² = 0, the answer will have one root X= 0.

Consider an equation of the form

3X² - 27 = 0

Dividing both sides by 3, we get the equation X² - 9 = 0, or it can be written X² = 9, the answer will have two roots X= 3 and X= -3.

Consider an equation of the form

2X² + 7 = 0

Dividing both sides by 2, we get the equation X² = -7/2. This equation has no real roots, since X² ≥ 0 for any real number X.

Consider an equation of the form

3X² + 5 X= 0

Factoring the left side of the equation, we get X(3X+ 5) = 0, the answer will have two roots X= 0, X=-5/3.

The most important thing when solving quadratic equations is to bring the quadratic equation to a standard form, memorize the formula for the roots of a general quadratic equation and not get confused in the signs.

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 \(D=b^2-4ac\)
\(D=(-4)^2-4\cdot1\cdot4=\) \(=16-16=0\)		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 : \(x=2\)

", 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 +

  1

  3

  = 0
• 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.

5x 2 − 14x + 17 = 0 −7x 2 − 13x + 8 = 0 −x 2 + x +

The equation	Odds
a = 5 b = −14 c = 17
a = −7 b = −13 c = 8

1

3

= 0

• a = −1
• b = 1
• c =

  1

  3

x 2 + 0.25x = 0

• a = 1
• b = 0.25
• c = 0

x 2 − 8 = 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:

• bring the quadratic equation to the general form “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 arises when the formula under the root turns out to be a negative number.

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

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

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

D = b 2 – 4ac.

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

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

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

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

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

D = 4 2 – 4 4 = 0

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

Answer: 2.

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

D = 1 2 – 4 2 3 = – 23

Answer: no roots.

Solve Equation 2 x 2 + 5x – 7 = 0.

D = 5 2 – 4 2 (–7) = 81

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

x 2 = (-5 + √81)/(2 2) = (-5 + 9)/4=1

Answer: – 3.5; 1.

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

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

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

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

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

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

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

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

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

Example. Solve the equation

3x 2 + 6x – 6 = 0.

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

D = 6 2 – 4 3 (– 6) = 36 + 72 = 108

√D = √108 = √(36 3) = 6√3

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

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

Answer: –1 – √3; –1 + √3

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

√(D 1) = √27 = √(9 3) = 3√3

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

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

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

D 2 = 2 2 – 4 (– 2) = 4 + 8 = 12

√(D 2) = √12 = √(4 3) = 2√3

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

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

Answer: –1 – √3; –1 + √3.

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

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

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

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

D = b 2 – 4ac.

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

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

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

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

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

D = 4 2 – 4 4 = 0

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

Answer: 2.

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

D = 1 2 – 4 2 3 = – 23

Answer: no roots.

Solve Equation 2 x 2 + 5x – 7 = 0.

D = 5 2 – 4 2 (–7) = 81

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

x 2 = (-5 + √81)/(2 2) = (-5 + 9)/4=1

Answer: – 3.5; 1.

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

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

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

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

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

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

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

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

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

Example. Solve the equation

3x 2 + 6x – 6 = 0.

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

D = 6 2 – 4 3 (– 6) = 36 + 72 = 108

√D = √108 = √(36 3) = 6√3

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

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

Answer: –1 – √3; –1 + √3

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

√(D 1) = √27 = √(9 3) = 3√3

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

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

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

D 2 = 2 2 – 4 (– 2) = 4 + 8 = 12

√(D 2) = √12 = √(4 3) = 2√3

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

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

Answer: –1 – √3; –1 + √3.

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

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