One of the most difficult topics for students is solving equations containing a variable under the modulus sign. Let's first figure out what this is connected with? Why, for example, do most children crack quadratic equations like nuts, but have so many problems with such a far from complex concept as a module?
In my opinion, all these difficulties are associated with the lack of clearly formulated rules for solving equations with a modulus. So, deciding quadratic equation, the student knows for sure that he needs to first apply the discriminant formula, and then the formulas for the roots of the quadratic equation. What to do if a modulus is found in the equation? We will try to clearly describe necessary plan actions in the case when the equation contains an unknown under the modulus sign. We will give several examples for each case.
But first, let's remember module definition. So, modulo the number a this number itself is called if a non-negative and -a, if number a less than zero. You can write it like this:
|a| = a if a ≥ 0 and |a| = -a if a< 0
Talking about geometric sense module, it should be remembered that each real number corresponds to a certain point on the number axis - its to 
coordinate. So, the module or absolute value of a number is the distance from this point to the origin of the numerical axis. The distance is always specified as a positive number. Thus, the modulus of any negative number is a positive number. By the way, even at this stage, many students begin to get confused. The module can contain any number, but the result of using the module is always a positive number.
Now let's move directly to solving the equations.
1. Consider an equation of the form |x| = c, where c is a real number. This equation can be solved using the modulus definition.
We divide all real numbers into three groups: those that Above zero, those that are less than zero, and the third group is the number 0. Let us write the solution in the form of a diagram:
(±c, if c > 0
If |x| = c, then x = (0, if c = 0
(no roots if with< 0
1) |x| = 5, because 5 > 0, then x = ±5;
2) |x| = -5, because -5< 0, то уравнение не имеет корней;
3) |x| = 0, then x = 0.
2. Equation of the form |f(x)| = b, where b > 0. To solve this equation it is necessary to get rid of the module. We do it this way: f(x) = b or f(x) = -b. Now you need to solve each of the resulting equations separately. If in the original equation b< 0, решений не будет.
1) |x + 2| = 4, because 4 > 0, then
x + 2 = 4 or x + 2 = -4
2) |x 2 – 5| = 11, because 11 > 0, then
x 2 – 5 = 11 or x 2 – 5 = -11
x 2 = 16 x 2 = -6
x = ± 4 no roots
3) |x 2 – 5x| = -8, because -8< 0, то уравнение не имеет корней.
3. An equation of the form |f(x)| = g(x). According to the meaning of the module, such an equation will have solutions if it right part greater than or equal to zero, i.e. g(x) ≥ 0. Then we will have:
f(x) = g(x) or f(x) = -g(x).
1) |2x – 1| = 5x – 10. This equation will have roots if 5x – 10 ≥ 0. This is where the solution of such equations begins.
1. O.D.Z. 5x – 10 ≥ 0
2. Solution:
2x – 1 = 5x – 10 or 2x – 1 = -(5x – 10)
3. We combine O.D.Z. and the solution, we get:
The root x = 11/7 does not fit the O.D.Z., it is less than 2, but x = 3 satisfies this condition.
Answer: x = 3
2) |x – 1| = 1 – x 2 .
1. O.D.Z. 1 – x 2 ≥ 0. Let’s solve this inequality using the interval method:
(1 – x)(1 + x) ≥ 0
2. Solution:
x – 1 = 1 – x 2 or x – 1 = -(1 – x 2)
x 2 + x – 2 = 0 x 2 – x = 0
x = -2 or x = 1 x = 0 or x = 1
3. We combine the solution and O.D.Z.:
Only roots x = 1 and x = 0 are suitable.
Answer: x = 0, x = 1.
4. Equation of the form |f(x)| = |g(x)|. Such an equation is equivalent to the following two equations f(x) = g(x) or f(x) = -g(x).
1) |x 2 – 5x + 7| = |2x – 5|. This equation is equivalent to the following two:
x 2 – 5x + 7 = 2x – 5 or x 2 – 5x +7 = -2x + 5
x 2 – 7x + 12 = 0 x 2 – 3x + 2 = 0
x = 3 or x = 4 x = 2 or x = 1
Answer: x = 1, x = 2, x = 3, x = 4.
5. Equations solved by the substitution method (variable replacement). This solution method is most easily explained in specific example. So, let us be given a quadratic equation with modulus:
x 2 – 6|x| + 5 = 0. By the modulus property x 2 = |x| 2, so the equation can be rewritten as follows:
|x| 2 – 6|x| + 5 = 0. Let's make the replacement |x| = t ≥ 0, then we will have:
t 2 – 6t + 5 = 0. Solving this equation, we find that t = 1 or t = 5. Let’s return to the replacement:
|x| = 1 or |x| = 5
x = ±1 x = ±5
Answer: x = -5, x = -1, x = 1, x = 5.
Let's look at another example:
x 2 + |x| – 2 = 0. By the modulus property x 2 = |x| 2, therefore
|x| 2 + |x| – 2 = 0. Let’s make the replacement |x| = t ≥ 0, then:
t 2 + t – 2 = 0. Solving this equation, we get t = -2 or t = 1. Let’s return to the replacement:
|x| = -2 or |x| = 1
No roots x = ± 1
Answer: x = -1, x = 1.
6. Another type of equations is equations with a “complex” modulus. Such equations include equations that have “modules within a module.” Equations of this type can be solved using the properties of the module.
1) |3 – |x|| = 4. We will act in the same way as in equations of the second type. Because 4 > 0, then we get two equations:
3 – |x| = 4 or 3 – |x| = -4.
Now let us express the modulus x in each equation, then |x| = -1 or |x| = 7.
We solve each of the resulting equations. There are no roots in the first equation, because -1< 0, а во втором x = ±7.
Answer x = -7, x = 7.
2) |3 + |x + 1|| = 5. We solve this equation in a similar way:
3 + |x + 1| = 5 or 3 + |x + 1| = -5
|x + 1| = 2 |x + 1| = -8
x + 1 = 2 or x + 1 = -2. No roots.
Answer: x = -3, x = 1.
There is also universal method solving equations with modulus. This is the interval method. But we will look at it later.
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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 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 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 into a 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 by 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 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: \()