The expression under the module must be. What is the modulus of a number in mathematics

This online math calculator will help you solve an equation or inequality with moduli. Program for solving equations and inequalities with moduli not only gives the answer to the problem, it leads detailed solution with explanations, i.e. displays the process of obtaining the result.

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A little theory.

Equations and inequalities with moduli

In a basic school algebra course, you may encounter the simplest equations and inequalities with moduli. To solve them, you can use a geometric method based on the fact that \(|x-a| \) is the distance on the number line between points x and a: \(|x-a| = \rho (x;\; a) \). For example, to solve the equation \(|x-3|=2\) you need to find points on the number line that are distant from point 3 at a distance of 2. There are two such points: \(x_1=1\) and \(x_2=5\) .

Solving the inequality \(|2x+7|

But the main way to solve equations and inequalities with moduli is associated with the so-called “revelation of the modulus by definition”:
if \(a \geq 0 \), then \(|a|=a \);
if \(a As a rule, an equation (inequality) with moduli is reduced to a set of equations (inequalities) that do not contain the modulus sign.

In addition to the above definition, the following statements are used:
1) If \(c > 0\), then the equation \(|f(x)|=c \) is equivalent to the set of equations: \(\left[\begin(array)(l) f(x)=c \\ f(x)=-c \end(array)\right. \)
2) If \(c > 0 \), then the inequality \(|f(x)| 3) If \(c \geq 0 \), then the inequality \(|f(x)| > c \) is equivalent to a set of inequalities : \(\left[\begin(array)(l) f(x) c \end(array)\right. \)
4) If both sides of the inequality \(f(x) EXAMPLE 1. Solve the equation \(x^2 +2|x-1| -6 = 0\).

If \(x-1 \geq 0\), then \(|x-1| = x-1\) and the given equation takes the form
\(x^2 +2(x-1) -6 = 0 \Rightarrow x^2 +2x -8 = 0 \).
If \(x-1 \(x^2 -2(x-1) -6 = 0 \Rightarrow x^2 -2x -4 = 0 \).
Thus, the given equation should be considered separately in each of the two indicated cases.
1) Let \(x-1 \geq 0 \), i.e. \(x\geq 1\). From the equation \(x^2 +2x -8 = 0\) we find \(x_1=2, \; x_2=-4\). The condition \(x \geq 1 \) is satisfied only by the value \(x_1=2\).
2) Let \(x-1 Answer: \(2; \;\; 1-\sqrt(5) \)

EXAMPLE 2. Solve the equation \(|x^2-6x+7| = \frac(5x-9)(3)\).

First way(module expansion by definition).
Reasoning as in example 1, we come to the conclusion that the given equation needs to be considered separately if two conditions are met: \(x^2-6x+7 \geq 0 \) or \(x^2-6x+7

1) If \(x^2-6x+7 \geq 0 \), then \(|x^2-6x+7| = x^2-6x+7 \) and the given equation takes the form \(x^2 -6x+7 = \frac(5x-9)(3) \Rightarrow 3x^2-23x+30=0 \). Having solved this quadratic equation, we get: \(x_1=6, \; x_2=\frac(5)(3) \).
Let's find out whether the value \(x_1=6\) satisfies the condition \(x^2-6x+7 \geq 0\). To do this, substitute the indicated value into the quadratic inequality. We get: \(6^2-6 \cdot 6+7 \geq 0 \), i.e. \(7 \geq 0 \) is a true inequality. This means that \(x_1=6\) is the root of the given equation.
Let's find out whether the value \(x_2=\frac(5)(3)\) satisfies the condition \(x^2-6x+7 \geq 0\). To do this, substitute the indicated value into the quadratic inequality. We get: \(\left(\frac(5)(3) \right)^2 -\frac(5)(3) \cdot 6 + 7 \geq 0 \), i.e. \(\frac(25)(9) -3 \geq 0 \) is an incorrect inequality. This means that \(x_2=\frac(5)(3)\) is not a root of the given equation.

2) If \(x^2-6x+7 Value \(x_3=3\) satisfies the condition \(x^2-6x+7 Value \(x_4=\frac(4)(3) \) does not satisfy the condition \ (x^2-6x+7 So, the given equation has two roots: \(x=6, \; x=3 \).

Second way. If the equation \(|f(x)| = h(x) \) is given, then with \(h(x) \(\left[\begin(array)(l) x^2-6x+7 = \frac (5x-9)(3) \\ x^2-6x+7 = -\frac(5x-9)(3) \end(array)\right. \)
Both of these equations were solved above (using the first method of solving the given equation), their roots are as follows: \(6,\; \frac(5)(3),\; 3,\; \frac(4)(3)\). The condition \(\frac(5x-9)(3) \geq 0 \) of these four values ​​is satisfied only by two: 6 and 3. This means that the given equation has two roots: \(x=6, \; x=3 \ ).

Third way(graphic).
1) Let's build a graph of the function \(y = |x^2-6x+7| \). First, let's construct a parabola \(y = x^2-6x+7\). We have \(x^2-6x+7 = (x-3)^2-2 \). The graph of the function \(y = (x-3)^2-2\) can be obtained from the graph of the function \(y = x^2 \) by shifting it by 3 scale units to the right (along the x-axis) and by 2 scale units down ( along the y-axis). The straight line x=3 is the axis of the parabola we are interested in. As control points for more accurate plotting, it is convenient to take point (3; -2) - the vertex of the parabola, point (0; 7) and point (6; 7) symmetrical to it relative to the axis of the parabola.
To now construct a graph of the function \(y = |x^2-6x+7| \), you need to leave unchanged those parts of the constructed parabola that lie not below the x-axis, and mirror that part of the parabola that lies below the x-axis relative to the x axis.
2) Let's build a graph of the linear function \(y = \frac(5x-9)(3)\). It is convenient to take points (0; –3) and (3; 2) as control points.

It is important that the point x = 1.8 of the intersection of the straight line with the abscissa axis is located to the right of the left point of intersection of the parabola with the abscissa axis - this is the point \(x=3-\sqrt(2) \) (since \(3-\sqrt(2 ) 3) Judging by the drawing, the graphs intersect at two points - A(3; 2) and B(6; 7).Substituting the abscissas of these points x = 3 and x = 6 into the given equation, we are convinced that both In another value, the correct numerical equality is obtained. This means that our hypothesis was confirmed - the equation has two roots: x = 3 and x = 6. Answer: 3; 6.

Comment. The graphical method, for all its elegance, is not very reliable. In the example considered, it worked only because the roots of the equation are integers.

EXAMPLE 3. Solve the equation \(|2x-4|+|x+3| = 8\)

First way
The expression 2x–4 becomes 0 at the point x = 2, and the expression x + 3 becomes 0 at the point x = –3. These two points divide the number line into three intervals: \(x

Consider the first interval: \((-\infty; \; -3) \).
If x Consider the second interval: \([-3; \; 2) \).
If \(-3 \leq x Consider the third interval: \()