Equation x 2 1. Solutions to the equation x2 = a. Incomplete quadratic equations

( (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 in the form 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.

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

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 equations in two variables.

Consider the equation 2x – y = 1. It turns into true equality for x = 2 and y = 3, so this pair of variable values ​​is a solution to the equation in question.

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

An equation with two unknowns can:

A) 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);

V) 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 equal to 3. The set of solutions given equation can be written in the form (k; 3 – k), where k is any real number.

The main methods for solving equations with two variables are methods based on factoring expressions, isolating a complete square, using the properties of a quadratic equation, limited expressions, and estimation methods. The equation is usually converted to a form from which a system for finding the unknowns can be obtained.

Factorization

Example 1.

Solve the equation: xy – 2 = 2x – y.

Solution.

We group the terms for the purpose of factorization:

(xy + y) – (2x + 2) = 0. From each bracket we take out a common factor:

y(x + 1) – 2(x + 1) = 0;

(x + 1)(y – 2) = 0. We have:

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

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

Equal to zero is not 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 bracket can be folded using the squared 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.

This means x = 2/3 and y = 3/2.

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

Estimation method

Example 3.

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

Solution.

In each bracket we select a complete square:

((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Let’s estimate the meaning of the expressions in parentheses.

(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, which means 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 consists of treating the equation 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 equation for 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, that is, 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 they indicate restrictions on variables.

Example 5.

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

Solution.

Let's rewrite the equation as x 2 = -5y 2 + 20x + 2. Right side 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 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 highlight the complete 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 provided |x| – 2 = 0 and y + 3 = 0. Thus, x = ± 2, y = -3.

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

Example 7.

For every pair of negative integers (x;y) satisfying the equation
x 2 – 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). Please indicate the smallest amount in your answer.

Solution.

Let's select complete 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. We get the sum of the squares of two integers equal to 37 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.

Don't despair if you have difficulty solving equations with two unknowns. With a little practice, you can handle 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 clear examples.
To produce solve quadratic equation online, first reduce the equation to general appearance:
ax 2 + bx + c = 0
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How to solve a quadratic equation

How to solve a quadratic equation:	Types of roots:
1. Reduce the quadratic equation to its general form: General view Аx 2 +Bx+C=0 Example: 3x - 2x 2 +1=-1 Reduce to -2x 2 +3x+2=0 2. Find the discriminant D. D=B 2 -4*A*C . For our example, D= 9-(4*(-2)*2)=9+16=25. 3. Finding the roots of the equation. x1=(-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 - even number, then it is more convenient to calculate the discriminant and roots using the formulas: D=К 2 -ac x1=(-K+D 1/2)/A x2=(-K-D 1/2)/A, Where K=B/2	1. Real roots. Moreover. x1 is not equal to x2 The situation occurs when D>0 and A is not equal to 0. 2. The real roots are the same. x1 equals x2 The situation occurs when D=0. However, neither A, nor B, nor C should be equal to 0. 3. Two complex roots. x1=d+ei, x2=d-ei, where i=-(1) 1/2 The situation occurs 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 countless 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 a few more illustrative examples of solutions to quadratic equations.

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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