Various methods for solving equations. Solve Quadratic Equation Online Check Equation Solution

In the 7th grade mathematics course, they first meet with equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why a number of problems fall out of sight, in which certain conditions are introduced on the coefficients of the equation that limit them. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although problems of this kind are encountered more and more often in the materials of the Unified State Examination and at entrance exams.

Which equation will be called an equation with two variables?

So, for example, the equations 5x + 2y = 10, x 2 + y 2 = 20, or xy = 12 are two-variable equations.

Consider the equation 2x - y = 1. It turns into a true equality at x = 2 and y = 3, so this pair of variable values is the solution to the equation under consideration.

Thus, the solution of any equation with two variables is the set of ordered pairs (x; y), the values of the variables that this equation turns into a true numerical equality.

An equation with two unknowns can:

but) have one solution. For example, the equation x 2 + 5y 2 = 0 has a unique solution (0; 0);

b) have multiple solutions. For example, (5 -|x|) 2 + (|y| – 2) 2 = 0 has 4 solutions: (5; 2), (-5; 2), (5; -2), (-5; - 2);

in) have no solutions. For example, the equation x 2 + y 2 + 1 = 0 has no solutions;

G) have infinitely many solutions. For example, x + y = 3. The solutions to this equation will be numbers whose sum is 3. The set of solutions to this equation can be written as (k; 3 - k), where k is any real number.

The main methods for solving equations with two variables are methods based on the decomposition of expressions into factors, the selection of a full square, the use of the properties of a quadratic equation, the boundedness of expressions, and evaluation methods. The equation, as a rule, is transformed into a form from which a system for finding unknowns can be obtained.

Factorization

Example 1

Solve the equation: xy - 2 = 2x - y.

Solution.

We group the terms for the purpose of factoring:

(xy + y) - (2x + 2) = 0. Take out the common factor from each bracket:

y(x + 1) – 2(x + 1) = 0;

(x + 1)(y - 2) = 0. We have:

y = 2, x is any real number or x = -1, y is any real number.

In this way, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.

Equality to zero of non-negative numbers

Example 2

Solve the equation: 9x 2 + 4y 2 + 13 = 12(x + y).

Solution.

Grouping:

(9x 2 - 12x + 4) + (4y 2 - 12y + 9) = 0. Now each parenthesis can be collapsed using the square difference formula.

(3x - 2) 2 + (2y - 3) 2 = 0.

The sum of two non-negative expressions is zero only if 3x - 2 = 0 and 2y - 3 = 0.

So x = 2/3 and y = 3/2.

Answer: (2/3; 3/2).

Evaluation Method

Example 3

Solve the equation: (x 2 + 2x + 2) (y 2 - 4y + 6) = 2.

Solution.

In each bracket, select the full square:

((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Estimate the meaning of the expressions in brackets.

(x + 1) 2 + 1 ≥ 1 and (y - 2) 2 + 2 ≥ 2, then the left side of the equation is always at least 2. Equality is possible if:

(x + 1) 2 + 1 = 1 and (y - 2) 2 + 2 = 2, so x = -1, y = 2.

Answer: (-1; 2).

Let's get acquainted with another method for solving equations with two variables of the second degree. This method is that the equation is considered as square with respect to some variable.

Example 4

Solve the equation: x 2 - 6x + y - 4√y + 13 = 0.

Solution.

Let's solve the equation as a quadratic one with respect to x. Let's find the discriminant:

D = 36 - 4(y - 4√y + 13) = -4y + 16√y - 16 = -4(√y - 2) 2 . The equation will have a solution only when D = 0, i.e., if y = 4. We substitute the value of y into the original equation and find that x = 3.

Answer: (3; 4).

Often in equations with two unknowns indicate restrictions on variables.

Example 5

Solve the equation in integers: x 2 + 5y 2 = 20x + 2.

Solution.

Let's rewrite the equation in the form x 2 = -5y 2 + 20x + 2. The right side of the resulting equation, when divided by 5, gives a remainder of 2. Therefore, x 2 is not divisible by 5. But the square of a number that is not divisible by 5 gives a remainder of 1 or 4. Thus equality is impossible and there are no solutions.

Answer: no roots.

Example 6

Solve the equation: (x 2 - 4|x| + 5) (y 2 + 6y + 12) = 3.

Solution.

Let's select the full squares in each bracket:

((|x| – 2) 2 + 1)((y + 3) 2 + 3) = 3. The left side of the equation is always greater than or equal to 3. Equality is possible if |x| – 2 = 0 and y + 3 = 0. Thus, x = ± 2, y = -3.

Answer: (2; -3) and (-2; -3).

Example 7

For each pair of negative integers (x; y) satisfying the equation
x 2 - 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). Answer the smallest amount.

Solution.

Select full squares:

(x 2 - 2xy + y 2) + (y 2 + 4y + 4) = 37;

(x - y) 2 + (y + 2) 2 = 37. Since x and y are integers, their squares are also integers. The sum of the squares of two integers, equal to 37, we get if we add 1 + 36. Therefore:

(x - y) 2 = 36 and (y + 2) 2 = 1

(x - y) 2 = 1 and (y + 2) 2 = 36.

Solving these systems and taking into account that x and y are negative, we find solutions: (-7; -1), (-9; -3), (-7; -8), (-9; -8).

Answer: -17.

Do not despair if you have difficulties when solving equations with two unknowns. With a little practice, you will be able to master any equation.

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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 understandable examples.
To produce solve quadratic equation online, first bring the equation to a general form:
ax2 + bx + c = 0
Fill in the form fields accordingly:

How to solve a quadratic equation

How to solve a quadratic equation:

Root types:

1. Bring the quadratic equation to a general form:
General view of Ax 2 +Bx+C=0
Example: 3x - 2x 2 +1=-1 Reduce to -2x 2 +3x+2=0

2. We find the discriminant D.
D=B 2 -4*A*C.
For our example, D= 9-(4*(-2)*2)=9+16=25.

3. We find the roots of the equation.
x1 \u003d (-B + D 1/2) / 2A.
For our case x1=(-3+5)/(-4)=-0.5
x2=(-B-D 1/2)/2A.
For our example x2=(-3-5)/(-4)=2
If B is an even number, then it is more convenient to calculate the discriminant and roots using the formulas:
D \u003d K 2 -ac
x1=(-K+D 1/2)/A
x2 \u003d (-K-D 1/2) / A,
Where K=B/2

1. Real roots. And. x1 is not equal to x2
The situation arises when D>0 and A is not equal to 0.

2. The real roots are the same. x1 equals x2
The situation arises when D=0. However, neither A, nor B, nor C must be equal to 0.

3. Two complex roots. x1=d+ei, x2=d-ei, where i=-(1) 1/2
The situation arises when D
4. The equation has one solution.
A=0, B and C are not equal to zero. The equation becomes linear.

5. The equation has an infinite number of solutions.
A=0, B=0, C=0.

6. The equation has no solutions.
A=0, B=0, C is not equal to 0.

To consolidate the algorithm, here are some more illustrative examples of solutions to quadratic equations.

Example 1. Solution of 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
the square root will be denoted as the number 1/2!
x1 \u003d (-B + D 1/2) / 2A \u003d (-3 + 7) / 2 \u003d 2
x2 \u003d (-B-D 1/2) / 2A \u003d (-3-7) / 2 \u003d -5

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

Example 2. Solving a quadratic equation with the same real roots.
x 2 - 8x + 16 = 0
A=1, B=-8, C=16
D \u003d k 2 - AC \u003d 16 - 16 \u003d 0
X=-k/A=4

Substitute
(x-4) * (x-4) \u003d (x-4) 2 \u003d X 2 - 8x + 16

Example 3. Solution of a quadratic equation with complex roots.
13x 2 - 4x + 1 = 0
A=1, B=-4, C=9
D \u003d b 2 - 4AC \u003d 16 - 4 * 13 * 1 \u003d 16 - 52 \u003d -36
The discriminant is negative - the roots are complex.

X1 \u003d (-B + D 1/2) / 2A \u003d (4 + 6i) / (2 * 13) \u003d 2/13 + 3i / 13
x2 \u003d (-B-D 1/2) / 2A \u003d (4-6i) / (2 * 13) \u003d 2 / 13-3i / 13
, where I is the square root of -1

Here are actually all possible cases of solving quadratic equations.
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Quadratic equations are studied in grade 8, so there is nothing complicated here. The ability to solve them is essential.

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 methods of solving, we note that all quadratic equations can be divided into three classes:

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

This is an important difference between quadratic and linear equations, 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 .

This formula must be known 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:

If D< 0, корней нет;
If D = 0, there is exactly one root;
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 think. Take a look at the examples and you will understand everything yourself:

A task. How many roots do quadratic equations have:

x 2 - 8x + 12 = 0;

5x2 + 3x + 7 = 0;

x 2 − 6x + 9 = 0.

We write 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 the same way:
a = 5; b = 3; c = 7;
D \u003d 3 2 - 4 5 7 \u003d 9 - 140 \u003d -131.

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

The discriminant is equal to zero - the root will be one.

Note that coefficients have been written out for each equation. Yes, it's long, yes, it's tedious - but you won't mix up the odds and don't make stupid mistakes. Choose for yourself: speed or quality.

By the way, if you “fill your hand”, after a while you will no longer need to write out 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 so many.

The roots of a quadratic equation

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

The basic formula for the roots of a quadratic equation

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

x 2 - 2x - 3 = 0;

15 - 2x - x2 = 0;

x2 + 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 be able to count, there will be no problems. Most often, errors occur when negative coefficients are substituted into the formula. Here, again, the technique described above will help: look at the formula literally, paint each step - and get rid of mistakes very soon.

Incomplete quadratic equations

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

x2 + 9x = 0;
x2 − 16 = 0.

It is easy to see that one of the terms is missing in these equations. Such quadratic equations are even easier to solve than standard ones: they do not even need to calculate 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 \u003d c \u003d 0. In this case, the equation takes the form ax 2 \u003d 0. Obviously, such an equation has a single root: x \u003d 0.

Let's consider other cases. Let b \u003d 0, then we get an incomplete quadratic equation of the form ax 2 + c \u003d 0. Let's slightly transform it:

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

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

As you can see, the 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 of 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 negative, there will be no roots at all.

Now let's deal with 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 factorize the polynomial:

Taking the common factor out of the bracket

The product is equal to zero when at least one of the factors is equal to zero. This is where the roots come from. In conclusion, we will analyze several of these equations:

A task. Solve quadratic equations:

x2 − 7x = 0;

5x2 + 30 = 0;

4x2 − 9 = 0.

x 2 − 7x = 0 ⇒ x (x − 7) = 0 ⇒ x 1 = 0; x2 = −(−7)/1 = 7.

5x2 + 30 = 0 ⇒ 5x2 = -30 ⇒ x2 = -6. There are no roots, because the 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 \u003d -1.5.

( (3 * x - 1) = 0;

-(3 * x - 1) = 0;

From here we get that there is one equation 3 * x - 1 = 0.

We got 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;
For 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 by the following formula x = - b/a.

From here we get that a \u003d 3, b \u003d - 1, which means that the equation has one root.

Checking Equation Solution

Substitute the found value x = 1/3 in the original expression |3 * x - 1| = 0, then we get:

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

In order to find the value of an expression, first, in turn, we calculate the multiplication or division, then the addition or subtraction operations are carried out. That is, we get:

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

|3 * x - 1| = 0;

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

1) 3 * x - 1 = 0;

We transfer the known values to one side, and the unknown ones 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;

We open the brackets. Since there is a minus sign in front of the brackets, when it is opened, 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.

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 of the cases 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 negative, the equation x^2=a, with 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 in the form x * x \u003d 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.

It follows from the definition of the square root 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, expand it.

(x+√a)*(x-√a)=0;

The product of two brackets is equal to zero if at least one of them is equal to zero. Consequently,

Hence, x1=√a x2=-√a.

This solution can be checked by plotting.

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.

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