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We offer you a convenient free online calculator for solving quadratic equations. You can quickly get and understand how they are solved using clear examples.
To produce solve quadratic equation online, first reduce the equation to general appearance:
ax 2 + bx + c = 0
Fill in the form fields accordingly:

How to solve a quadratic equation

Example 1. Solving an ordinary quadratic equation with different real roots.
x 2 + 3x -10 = 0
In this equation
A=1, B=3, C=-10
D=B 2 -4*A*C = 9-4*1*(-10) = 9+40 = 49
Square root We will denote it as the number 1/2!
x1=(-B+D 1/2)/2A = (-3+7)/2 = 2
x2=(-B-D 1/2)/2A = (-3-7)/2 = -5

To check, let's substitute:
(x-2)*(x+5) = x2 -2x +5x – 10 = x2 + 3x -10

Example 2. Solving a quadratic equation with matching real roots.
x 2 – 8x + 16 = 0
A=1, B = -8, C=16
D = k 2 – AC = 16 – 16 = 0
X = -k/A = 4

Let's substitute
(x-4)*(x-4) = (x-4)2 = X 2 – 8x + 16

Example 3. Solving a quadratic equation with complex roots.
13x 2 – 4x + 1 = 0
A=1, B = -4, C=9
D = b 2 – 4AC = 16 – 4*13*1 = 16 - 52 = -36
The discriminant is negative – the roots are complex.

X1=(-B+D 1/2)/2A = (4+6i)/(2*13) = 2/13+3i/13
x2=(-B-D 1/2)/2A = (4-6i)/(2*13) = 2/13-3i/13
, where I is the square root of -1

Here are actually all the possible cases of solving quadratic equations.
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( (3 * x – 1) = 0;

-(3 * x – 1) = 0;

From here we see that there is one equation 3 * x – 1 = 0.

We received a linear equation in the form 3 * x – 1 = 0

In order to solve the equation, we determine what properties the equation has:

• The equation is linear, and is written as a * x + b = 0, where a and b are any numbers;
• When a = b = 0, the equation has an infinite number of solutions;
• If a = 0, b ≠ 0, the equation has no solution;
• If a ≠ 0, b = 0, the equation has a solution: x = 0;
• If a and b are any numbers other than 0, then the root is found using the following formula x = - b/a.

From here we get that a = 3, b = - 1, which means the equation has one root.

Checking the solution to the equation

Let's substitute the found value x = 1/3 into the original expression |3 * x - 1| = 0, then we get:

|3 * 1/3 - 1| = 0;

In order to find the value of an expression, we first calculate multiplication or division in turn, then add or subtract. That is, we get:

This means x = 1/3 is the root of the equation |3 * x - 1| = 0.

|3 * x - 1| = 0;

The module opens with a plus and minus sign. We get 2 equations:

1) 3 * x - 1 = 0;

We transfer known values ​​to one side, and unknown values ​​to the other side. When transferring values, their signs change to the opposite sign. That is, we get:
3 * x = 0 + 1;
3 * x = 1;
x = 1/3;

2) - (3 * x - 1) = 0;

Opening the parentheses. Since there is a minus sign in front of the parentheses, when they are expanded, the signs of the values ​​change to the opposite sign. That is, we get:
- 3 * x + 1 = 0;
- 3 * x = - 1;
x = - 1/(- 3);
x = 1/3;
Answer: x = 1/3.

Let's consider the equation x^2=a, where a can be an arbitrary number. There are three cases of solving this equation, depending on the value that the number a (a0) takes.

Let's consider each case separately.

Examples of different cases of the equation x^2=a

x^2=a, for a<0

Since the square of any real number cannot be a negative number, the equation x^2=a, for a

x^2=a, with a=0

In this case, the equation has one root. This root is the number 0. Since the equation can be rewritten as x*x=0, it is also sometimes said that this equation has two roots that are equal to each other and equal to 0.

x^2=a, for a>0

In this case, the equation x^2=a, for a, it is solved as follows. First we move a to the left side.

From the definition of a square root it follows that a can be written in the following form: a=(√a)^2. Then the equation can be rewritten as follows:

x^2 - (√a)^2 = 0.

On the left side we see the formula for the difference of squares; let’s expand it.

(x+√a)*(x-√a)=0;

The product of two parentheses is equal to zero if at least one of them is equal to zero. Hence,

Hence, x1=√a x2=-√a.

This solution can be checked by plotting a graph.

For example, let's do this for the equation x^2 = 4.

To do this, you need to build two graphs y=x^2 and y=4. And look at the x coordinates of their intersection points. The roots should be 2 and -2. Everything is clearly visible in the figure.

Need help with your studies?

I. Linear equations

II. Quadratic equations

ax 2 + bx +c= 0, a≠ 0, otherwise the equation becomes linear

The roots of a quadratic equation can be calculated different ways, For example:

We are good at solving quadratic equations. Many equations of higher degrees can be reduced to quadratic equations.

III.

Equations reduced to quadratic. ax change of variable: a) biquadratic equation bx 2n+ c = 0,a ≠ 0,n+ ≥ 2

n

2) symmetric equation of degree 3 – equation of the form

ax 4 + bx 3 + 3) symmetric equation of degree 4 – equation of the form 2 +cx + bx = 0, bx a ≠ 0, coefficients a b c b a

ax 4 + bx 3 + 3) symmetric equation of degree 4 – equation of the form 2 –cx + bx = 0, bx or ≠ 0, coefficients

a b c (–b) a Because x Because= 0 is not a root of the equation, then it is possible to divide both sides of the equation by

2, then we get: . bx(By making the substitution we solve the quadratic equation 2 – 2) + t + bt = 0

c Because 4 – 2Because 3 – Because 2 – 2Because For example, let's solve the equation Because 2 ,

+ 1 = 0, divide both sides by By making the substitution we solve the quadratic equation 2 – 2By making the substitution we solve the quadratic equation – 3 = 0

, after replacement we get the equation

– the equation has no roots. 4) Equation of the form ()(x–a)(x–b)(x–c) = x–d Ax 2 , coefficients

ab = cd For example, ()(x+2)(x +3)(x+8) = x+12 4x Because 2 + 14Because+ 24)(Because 2 +11Because + 24) = 4Because 2. Multiplying 1–4 and 2–3 brackets, we get ( Because 2, divide both sides of the equation by

2, we get: By making the substitution we solve the quadratic equation+ 14)(By making the substitution we solve the quadratic equation + 11) = 4.

We have (

5) Homogeneous equation of degree 2 - an equation of the form P(x,y) = 0, where P(x,y) is a polynomial, each term of which has degree 2.

Answer: -2; -0.5; 0

IV. All the above equations are recognizable and typical, but what about equations of arbitrary form? Let a polynomial be given P Because) = bx n ( Because n bx n+ Because n-1 bx n-1 + ...+ 1x+ a bx 0 , where

n ≠ 0

Let's consider the method of reducing the degree of the equation. bx It is known that if the coefficients bx are integers and Let a polynomial be given P Because n = 1, then the integer roots of the equation bx) = 0 are among the divisors of the free term Because 4 + 2Because 3 – 2Because 2 – 6Because 0 . For example, Let a polynomial be given+ 5 = 0, divisors of the number 5 are the numbers 5; -5; 1; -1. Then Because 4 (1) = 0, i.e. Let a polynomial be given 4 (Because) = 0 by dividing the polynomial with a “corner” by the factor x –1, we obtain

Let a polynomial be given 4 (Because) = (Because – 1)(Because 3 + 3Because 2 + Because – 5).

Likewise, Let a polynomial be given 3 (1) = 0, then Let a polynomial be given 4 (Because) = (Because – 1)(Because – 1)(Because 2 + 4Because+5), i.e. the equation Let a polynomial be given 4 (x) = 0 has roots Because 1 = Because 2 = 1. Let's show more short solution this equation (using Horner's scheme).

Means, Because 1 = 1 means Because 2 = 1.

So, ( Because– 1) 2 (Because 2 + 4Because + 5) = 0

What did we do? We lowered the degree of the equation.

V. Consider symmetric equations of degree 3 and 5.

A) ax 3 + bx 2 + bx + bx= 0, obviously Because= –1 is the root of the equation, then we lower the degree of the equation to two.

b) ax 5 + bx 4 + 3) symmetric equation of degree 4 – equation of the form 3 + 3) symmetric equation of degree 4 – equation of the form 2 + bx + bx= 0, obviously Because= –1 is the root of the equation, then we lower the degree of the equation to two.

For example, let's show the solution to equation 2 Because 5 + 3Because 4 – 5Because 3 – 5Because 2 + 3Because + = 0

Because = –1

We get ( Because – 1) 2 (Because + 1)(2Because 2 + 5Because+ 2) = 0. This means that the roots of the equation are: 1; 1; -1; –2; –0.5.

VI. Here is a list of different equations to solve in class and at home.

I suggest the reader solve equations 1–7 himself and get the answers...

Quadratic equations are studied in 8th grade, so there is nothing complicated here. The ability to solve them is absolutely necessary.

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a, b and c are arbitrary numbers, and a ≠ 0.

Before studying specific solution methods, note that all quadratic equations can be divided into three classes:

1. Have no roots;
2. Have exactly one root;
3. They have two different roots.

This is an important difference between quadratic equations and linear ones, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.

Discriminant

Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac.

You need to know this formula by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant you can determine how many roots a quadratic equation has. Namely:

1. If D< 0, корней нет;
2. If D = 0, there is exactly one root;
3. If D > 0, there will be two roots.

Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people believe. Take a look at the examples and you will understand everything yourself:

Task. How many roots do quadratic equations have:

1. x 2 − 8x + 12 = 0;
2. 5x 2 + 3x + 7 = 0;
3. x 2 − 6x + 9 = 0.

Let's write out the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16

So the discriminant is positive, so the equation has two different roots. We analyze the second equation in a similar way:
a = 5; b = 3; c = 7;
D = 3 2 − 4 5 7 = 9 − 140 = −131.

The discriminant is negative, there are no roots. The last equation left is:
a = 1; b = −6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.

The discriminant is zero - the root will be one.

Please note that coefficients have been written down for each equation. Yes, it’s long, yes, it’s tedious, but you won’t mix up the odds and make stupid mistakes. Choose for yourself: speed or quality.

By the way, if you get the hang of it, after a while you won’t need to write down all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not that much.

Roots of a quadratic equation

Now let's move on to the solution itself. If the discriminant D > 0, the roots can be found using the formulas:

Basic formula for the roots of a quadratic equation

When D = 0, you can use any of these formulas - you will get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.

1. x 2 − 2x − 3 = 0;
2. 15 − 2x − x 2 = 0;
3. x 2 + 12x + 36 = 0.

First equation:
x 2 − 2x − 3 = 0 ⇒ a = 1; b = −2; c = −3;
D = (−2) 2 − 4 1 (−3) = 16.

D > 0 ⇒ the equation has two roots. Let's find them:

Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 · (−1) · 15 = 64.

D > 0 ⇒ the equation again has two roots. Let's find them

\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]

Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.

D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:

As you can see from the examples, everything is very simple. If you know the formulas and can count, there will be no problems. Most often, errors occur when substituting negative coefficients into the formula. Here again, the technique described above will help: look at the formula literally, write down each step - and very soon you will get rid of mistakes.

Incomplete quadratic equations

It happens that a quadratic equation is slightly different from what is given in the definition. For example:

1. x 2 + 9x = 0;
2. x 2 − 16 = 0.

It is easy to notice that these equations are missing one of the terms. Such quadratic equations are even easier to solve than standard ones: they don’t even require calculating the discriminant. So, let's introduce a new concept:

The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.

Of course, a very difficult case is possible when both of these coefficients are equal to zero: b = c = 0. In this case, the equation takes the form ax 2 = 0. Obviously, such an equation has a single root: x = 0.

Let's consider the remaining cases. Let b = 0, then we obtain an incomplete quadratic equation of the form ax 2 + c = 0. Let us transform it a little:

Since the arithmetic square root exists only of a non-negative number, the last equality makes sense only for (−c /a) ≥ 0. Conclusion:

1. If in an incomplete quadratic equation of the form ax 2 + c = 0 the inequality (−c /a) ≥ 0 is satisfied, there will be two roots. The formula is given above;
2. If (−c /a)< 0, корней нет.

As you can see, a discriminant was not required—there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c /a) ≥ 0. It is enough to express the value x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If it is negative, there will be no roots at all.

Now let's look at equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factor the polynomial:

The product is zero when at least one of the factors is zero. This is where the roots come from. In conclusion, let’s look at a few of these equations:

Task. Solve quadratic equations:

1. x 2 − 7x = 0;
2. 5x 2 + 30 = 0;
3. 4x 2 − 9 = 0.

x 2 − 7x = 0 ⇒ x · (x − 7) = 0 ⇒ x 1 = 0; x 2 = −(−7)/1 = 7.

5x 2 + 30 = 0 ⇒ 5x 2 = −30 ⇒ x 2 = −6. There are no roots, because a square cannot be equal to a negative number.

4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 = −1.5.