Mister Exam
|x^2-2x|+|x+1|>5 inequation
The solution
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| 2 | |x - 2*x| + |x + 1| > 5
$$\left|{x + 1}\right| + \left|{x^{2} - 2 x}\right| > 5$$
|x + 1| + |x^2 - 2*x| > 5
Detail solution
Given the inequality:
$$\left|{x + 1}\right| + \left|{x^{2} - 2 x}\right| > 5$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{x + 1}\right| + \left|{x^{2} - 2 x}\right| = 5$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x + 1 \geq 0$$
$$x^{2} - 2 x \geq 0$$
or
$$\left(-1 \leq x \wedge x \leq 0\right) \vee \left(2 \leq x \wedge x < \infty\right)$$
we get the equation
$$\left(x + 1\right) + \left(x^{2} - 2 x\right) - 5 = 0$$
after simplifying we get
$$x^{2} - x - 4 = 0$$
the solution in this interval:
$$x_{1} = \frac{1}{2} - \frac{\sqrt{17}}{2}$$
but x1 not in the inequality interval
$$x_{2} = \frac{1}{2} + \frac{\sqrt{17}}{2}$$
2.
$$x + 1 \geq 0$$
$$x^{2} - 2 x < 0$$
or
$$0 < x \wedge x < 2$$
we get the equation
$$\left(x + 1\right) + \left(- x^{2} + 2 x\right) - 5 = 0$$
after simplifying we get
$$- x^{2} + 3 x - 4 = 0$$
the solution in this interval:
$$x_{3} = \frac{3}{2} - \frac{\sqrt{7} i}{2}$$
but x3 not in the inequality interval
$$x_{4} = \frac{3}{2} + \frac{\sqrt{7} i}{2}$$
but x4 not in the inequality interval
3.
$$x + 1 < 0$$
$$x^{2} - 2 x \geq 0$$
or
$$-\infty < x \wedge x < -1$$
we get the equation
$$\left(- x - 1\right) + \left(x^{2} - 2 x\right) - 5 = 0$$
after simplifying we get
$$x^{2} - 3 x - 6 = 0$$
the solution in this interval:
$$x_{5} = \frac{3}{2} - \frac{\sqrt{33}}{2}$$
$$x_{6} = \frac{3}{2} + \frac{\sqrt{33}}{2}$$
but x6 not in the inequality interval
4.
$$x + 1 < 0$$
$$x^{2} - 2 x < 0$$
The inequality system has no solutions, see the next condition
$$x_{1} = \frac{1}{2} + \frac{\sqrt{17}}{2}$$
$$x_{2} = \frac{3}{2} - \frac{\sqrt{33}}{2}$$
$$x_{1} = \frac{1}{2} + \frac{\sqrt{17}}{2}$$
$$x_{2} = \frac{3}{2} - \frac{\sqrt{33}}{2}$$
This roots
$$x_{2} = \frac{3}{2} - \frac{\sqrt{33}}{2}$$
$$x_{1} = \frac{1}{2} + \frac{\sqrt{17}}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$\left(\frac{3}{2} - \frac{\sqrt{33}}{2}\right) + - \frac{1}{10}$$
=
$$\frac{7}{5} - \frac{\sqrt{33}}{2}$$
substitute to the expression
$$\left|{x + 1}\right| + \left|{x^{2} - 2 x}\right| > 5$$
$$\left|{\left(\frac{7}{5} - \frac{\sqrt{33}}{2}\right) + 1}\right| + \left|{\left(\frac{7}{5} - \frac{\sqrt{33}}{2}\right)^{2} - 2 \left(\frac{7}{5} - \frac{\sqrt{33}}{2}\right)}\right| > 5$$
one of the solutions of our inequality is:
$$x < \frac{3}{2} - \frac{\sqrt{33}}{2}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{3}{2} - \frac{\sqrt{33}}{2}$$
$$x > \frac{1}{2} + \frac{\sqrt{17}}{2}$$
$$\left|{x + 1}\right| + \left|{x^{2} - 2 x}\right| > 5$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{x + 1}\right| + \left|{x^{2} - 2 x}\right| = 5$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x + 1 \geq 0$$
$$x^{2} - 2 x \geq 0$$
or
$$\left(-1 \leq x \wedge x \leq 0\right) \vee \left(2 \leq x \wedge x < \infty\right)$$
we get the equation
$$\left(x + 1\right) + \left(x^{2} - 2 x\right) - 5 = 0$$
after simplifying we get
$$x^{2} - x - 4 = 0$$
the solution in this interval:
$$x_{1} = \frac{1}{2} - \frac{\sqrt{17}}{2}$$
but x1 not in the inequality interval
$$x_{2} = \frac{1}{2} + \frac{\sqrt{17}}{2}$$
2.
$$x + 1 \geq 0$$
$$x^{2} - 2 x < 0$$
or
$$0 < x \wedge x < 2$$
we get the equation
$$\left(x + 1\right) + \left(- x^{2} + 2 x\right) - 5 = 0$$
after simplifying we get
$$- x^{2} + 3 x - 4 = 0$$
the solution in this interval:
$$x_{3} = \frac{3}{2} - \frac{\sqrt{7} i}{2}$$
but x3 not in the inequality interval
$$x_{4} = \frac{3}{2} + \frac{\sqrt{7} i}{2}$$
but x4 not in the inequality interval
3.
$$x + 1 < 0$$
$$x^{2} - 2 x \geq 0$$
or
$$-\infty < x \wedge x < -1$$
we get the equation
$$\left(- x - 1\right) + \left(x^{2} - 2 x\right) - 5 = 0$$
after simplifying we get
$$x^{2} - 3 x - 6 = 0$$
the solution in this interval:
$$x_{5} = \frac{3}{2} - \frac{\sqrt{33}}{2}$$
$$x_{6} = \frac{3}{2} + \frac{\sqrt{33}}{2}$$
but x6 not in the inequality interval
4.
$$x + 1 < 0$$
$$x^{2} - 2 x < 0$$
The inequality system has no solutions, see the next condition
$$x_{1} = \frac{1}{2} + \frac{\sqrt{17}}{2}$$
$$x_{2} = \frac{3}{2} - \frac{\sqrt{33}}{2}$$
$$x_{1} = \frac{1}{2} + \frac{\sqrt{17}}{2}$$
$$x_{2} = \frac{3}{2} - \frac{\sqrt{33}}{2}$$
This roots
$$x_{2} = \frac{3}{2} - \frac{\sqrt{33}}{2}$$
$$x_{1} = \frac{1}{2} + \frac{\sqrt{17}}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$\left(\frac{3}{2} - \frac{\sqrt{33}}{2}\right) + - \frac{1}{10}$$
=
$$\frac{7}{5} - \frac{\sqrt{33}}{2}$$
substitute to the expression
$$\left|{x + 1}\right| + \left|{x^{2} - 2 x}\right| > 5$$
$$\left|{\left(\frac{7}{5} - \frac{\sqrt{33}}{2}\right) + 1}\right| + \left|{\left(\frac{7}{5} - \frac{\sqrt{33}}{2}\right)^{2} - 2 \left(\frac{7}{5} - \frac{\sqrt{33}}{2}\right)}\right| > 5$$
2
/ ____\ ____
26 |7 \/ 33 | 3*\/ 33 > 5
- -- + |- - ------| + --------
5 \5 2 / 2 one of the solutions of our inequality is:
$$x < \frac{3}{2} - \frac{\sqrt{33}}{2}$$
_____ _____
\ /
-------ο-------ο-------
x2 x1Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{3}{2} - \frac{\sqrt{33}}{2}$$
$$x > \frac{1}{2} + \frac{\sqrt{17}}{2}$$
Solving inequality on a graph
Rapid solution
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/ / ____\ / ____ \\ | | 3 \/ 33 | | 1 \/ 17 || Or|And|-oo < x, x < - - ------|, And|x < oo, - + ------ < x|| \ \ 2 2 / \ 2 2 //
$$\left(-\infty < x \wedge x < \frac{3}{2} - \frac{\sqrt{33}}{2}\right) \vee \left(x < \infty \wedge \frac{1}{2} + \frac{\sqrt{17}}{2} < x\right)$$
((-oo < x)∧(x < 3/2 - sqrt(33)/2))∨((x < oo)∧(1/2 + sqrt(17)/2 < x))
Rapid solution 2
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____ ____
3 \/ 33 1 \/ 17
(-oo, - - ------) U (- + ------, oo)
2 2 2 2
$$x\ in\ \left(-\infty, \frac{3}{2} - \frac{\sqrt{33}}{2}\right) \cup \left(\frac{1}{2} + \frac{\sqrt{17}}{2}, \infty\right)$$
x in Union(Interval.open(-oo, 3/2 - sqrt(33)/2), Interval.open(1/2 + sqrt(17)/2, oo))