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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|x| or abs(x) - module xEnter an equation or inequality with moduli
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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 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.
Except 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 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, let's substitute specified value V 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: \(6,\; \frac(5)(3),\; 3,\; \frac(4)(3)\). Condition \(\frac(5x-9)(3) \geq 0 \) from these four values satisfy only 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 3 scale units to the right (along the x-axis) and 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 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 in both cases. 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;
Comment. Graphic method for all its elegance, it 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: \() - they will automatically count this as an incorrect answer. Also during testing, if not specified strict inequality with modules, then look for areas with square brackets among the solutions.
On the interval (-3;0), expanding the module, we change the sign of the function to the opposite one 
Taking into account the area of inequality disclosure, the solution will have the form
Together with the previous area this will give two half-intervals
Example 5. Find a solution to the inequality
9x^2-|x-3|>=9x-2 
Solution:
A non-strict inequality is given whose submodular function is equal to zero at the point x=3. For smaller values it is negative, for larger values it is positive. Expand the module on the interval x<3.
Finding the discriminant of the equation
and roots 
Substituting point zero, we find out that on the interval [-1/9;1] the quadratic function is negative, therefore the interval is a solution. Next we expand the module at x>3 
There are several ways to solve inequalities containing a modulus. Let's look at some of them.
1) Solving the inequality using the geometric property of the module.
Let me remind you what it is geometric property modulus: the modulus of a number x is the distance from the origin to the point with coordinate x.
When solving inequalities using this method, two cases may arise:
1. |x| ≤ b, 
And the inequality with modulus obviously reduces to a system of two inequalities. Here the sign can be strict, in which case the dots in the picture will be “punctured”.
2. |x| ≥ b, then the solution picture looks like this:
And the inequality with modulus obviously reduces to a combination of two inequalities. Here the sign can be strict, in which case the dots in the picture will be “punctured”.
Example 1.
Solve the inequality |4 – |x|| ≥ 3.
Solution.
This inequality is equivalent to the following set:
U [-1;1] U
Example 2.
Solve the inequality ||x+2| – 3| ≤ 2.
Solution.
This inequality is equivalent to the following system.
(|x + 2| – 3 ≥ -2
(|x + 2| – 3 ≤ 2,
(|x + 2| ≥ 1
(|x + 2| ≤ 5.
Let us solve separately the first inequality of the system. It is equivalent to the following set:
U[-1; 3].
2) Solving inequalities using the definition of the modulus.
Let me remind you first module definition.
|a| = a if a ≥ 0 and |a| = -a if a< 0.
For example, |34| = 34, |-21| = -(-21) = 21.
Example 1.
Solve the inequality 3|x – 1| ≤ x+3.
Solution.
Using the module definition we get two systems:
(x – 1 ≥ 0
(3(x – 1) ≤ x + 3
(x – 1< 0
(-3(x – 1) ≤ x + 3.
Solving the first and second systems separately, we obtain:
(x ≥ 1
(x ≤ 3,
(x< 1
(x ≥ 0.
The solution to the original inequality will be all solutions of the first system and all solutions of the second system.
Answer: x € .
3) Solving inequalities by squaring.
Example 1.
Solve the inequality |x 2 – 1|< | x 2 – x + 1|.
Solution.
Let us square both sides of the inequality. Let me note that you can square both sides of the inequality only if they are both positive. IN in this case We have modules on both the left and right, so we can do this.
(|x 2 – 1|) 2< (|x 2 – x + 1|) 2 .
Now let's use the following property of the module: (|x|) 2 = x 2 .
(x 2 – 1) 2< (x 2 – x + 1) 2 ,
(x 2 – 1) 2 – (x 2 – x + 1) 2< 0.
(x 2 – 1 – x 2 + x – 1)(x 2 – 1 + x 2 – x + 1)< 0,
(x – 2)(2x 2 – x)< 0,
x(x – 2)(2x – 1)< 0.
We solve using the interval method.
Answer: x € (-∞; 0) U (1/2; 2)
4) Solving inequalities by changing variables.
Example.
Solve the inequality (2x + 3) 2 – |2x + 3| ≤ 30.
Solution.
Note that (2x + 3) 2 = (|2x + 3|) 2 . Then we get the inequality
(|2x + 3|) 2 – |2x + 3| ≤ 30.
Let's make the change y = |2x + 3|.
Let's rewrite our inequality taking into account the replacement.
y 2 – y ≤ 30,
y 2 – y – 30 ≤ 0.
Let's decompose quadratic trinomial, standing on the left, into factors.
y1 = (1 + 11) / 2,
y2 = (1 – 11) / 2,
(y – 6)(y + 5) ≤ 0.
Let's solve using the interval method and get:
Let's go back to the replacement:
5 ≤ |2x + 3| ≤ 6.
This double inequality is equivalent to the system of inequalities:
(|2x + 3| ≤ 6
(|2x + 3| ≥ -5.
Let's solve each of the inequalities separately.
The first is equivalent to the system
(2x + 3 ≤ 6
(2x + 3 ≥ -6.
Let's solve it.
(x ≤ 1.5
(x ≥ -4.5.
The second inequality obviously holds for all x, since the modulus is, by definition, a positive number. Since the solution to the system is all x that simultaneously satisfy both the first and second inequality of the system, then the solution to the original system will be the solution to its first double inequality (after all, the second is true for all x).
Answer: x € [-4.5; 1.5].
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