Mister Exam
x^8-x^6-4x^4++x^2+1>=0 inequation
The solution
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8 6 4 2 x - x - 4*x + x + 1 >= 0
$$\left(x^{2} + \left(- 4 x^{4} + \left(x^{8} - x^{6}\right)\right)\right) + 1 \geq 0$$
x^2 - 4*x^4 + x^8 - x^6 + 1 >= 0
Detail solution
Given the inequality:
$$\left(x^{2} + \left(- 4 x^{4} + \left(x^{8} - x^{6}\right)\right)\right) + 1 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x^{2} + \left(- 4 x^{4} + \left(x^{8} - x^{6}\right)\right)\right) + 1 = 0$$
Solve:
$$x_{1} = - \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{2} = \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{3} = - \sqrt{1 - \sqrt{2}}$$
$$x_{4} = \sqrt{1 - \sqrt{2}}$$
$$x_{5} = - \sqrt{1 + \sqrt{2}}$$
$$x_{6} = \sqrt{1 + \sqrt{2}}$$
$$x_{7} = - \sqrt{- \frac{\sqrt{5}}{2} - \frac{1}{2}}$$
$$x_{8} = \sqrt{- \frac{\sqrt{5}}{2} - \frac{1}{2}}$$
Exclude the complex solutions:
$$x_{1} = - \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{2} = \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{3} = - \sqrt{1 + \sqrt{2}}$$
$$x_{4} = \sqrt{1 + \sqrt{2}}$$
This roots
$$x_{3} = - \sqrt{1 + \sqrt{2}}$$
$$x_{1} = - \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{2} = \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{4} = \sqrt{1 + \sqrt{2}}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{3}$$
For example, let's take the point
$$x_{0} = x_{3} - \frac{1}{10}$$
=
$$- \sqrt{1 + \sqrt{2}} - \frac{1}{10}$$
=
$$- \sqrt{1 + \sqrt{2}} - \frac{1}{10}$$
substitute to the expression
$$\left(x^{2} + \left(- 4 x^{4} + \left(x^{8} - x^{6}\right)\right)\right) + 1 \geq 0$$
$$1 + \left(\left(- \sqrt{1 + \sqrt{2}} - \frac{1}{10}\right)^{2} + \left(- 4 \left(- \sqrt{1 + \sqrt{2}} - \frac{1}{10}\right)^{4} + \left(- \left(- \sqrt{1 + \sqrt{2}} - \frac{1}{10}\right)^{6} + \left(- \sqrt{1 + \sqrt{2}} - \frac{1}{10}\right)^{8}\right)\right)\right) \geq 0$$
one of the solutions of our inequality is:
$$x \leq - \sqrt{1 + \sqrt{2}}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq - \sqrt{1 + \sqrt{2}}$$
$$x \geq - \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}} \wedge x \leq \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x \geq \sqrt{1 + \sqrt{2}}$$
$$\left(x^{2} + \left(- 4 x^{4} + \left(x^{8} - x^{6}\right)\right)\right) + 1 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x^{2} + \left(- 4 x^{4} + \left(x^{8} - x^{6}\right)\right)\right) + 1 = 0$$
Solve:
$$x_{1} = - \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{2} = \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{3} = - \sqrt{1 - \sqrt{2}}$$
$$x_{4} = \sqrt{1 - \sqrt{2}}$$
$$x_{5} = - \sqrt{1 + \sqrt{2}}$$
$$x_{6} = \sqrt{1 + \sqrt{2}}$$
$$x_{7} = - \sqrt{- \frac{\sqrt{5}}{2} - \frac{1}{2}}$$
$$x_{8} = \sqrt{- \frac{\sqrt{5}}{2} - \frac{1}{2}}$$
Exclude the complex solutions:
$$x_{1} = - \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{2} = \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{3} = - \sqrt{1 + \sqrt{2}}$$
$$x_{4} = \sqrt{1 + \sqrt{2}}$$
This roots
$$x_{3} = - \sqrt{1 + \sqrt{2}}$$
$$x_{1} = - \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{2} = \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x_{4} = \sqrt{1 + \sqrt{2}}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{3}$$
For example, let's take the point
$$x_{0} = x_{3} - \frac{1}{10}$$
=
$$- \sqrt{1 + \sqrt{2}} - \frac{1}{10}$$
=
$$- \sqrt{1 + \sqrt{2}} - \frac{1}{10}$$
substitute to the expression
$$\left(x^{2} + \left(- 4 x^{4} + \left(x^{8} - x^{6}\right)\right)\right) + 1 \geq 0$$
$$1 + \left(\left(- \sqrt{1 + \sqrt{2}} - \frac{1}{10}\right)^{2} + \left(- 4 \left(- \sqrt{1 + \sqrt{2}} - \frac{1}{10}\right)^{4} + \left(- \left(- \sqrt{1 + \sqrt{2}} - \frac{1}{10}\right)^{6} + \left(- \sqrt{1 + \sqrt{2}} - \frac{1}{10}\right)^{8}\right)\right)\right) \geq 0$$
2 8 6 4
/ ___________\ / ___________\ / ___________\ / ___________\
| 1 / ___ | | 1 / ___ | | 1 / ___ | | 1 / ___ | >= 0
1 + |- -- - \/ 1 + \/ 2 | + |- -- - \/ 1 + \/ 2 | - |- -- - \/ 1 + \/ 2 | - 4*|- -- - \/ 1 + \/ 2 |
\ 10 / \ 10 / \ 10 / \ 10 / one of the solutions of our inequality is:
$$x \leq - \sqrt{1 + \sqrt{2}}$$
_____ _____ _____
\ / \ /
-------•-------•-------•-------•-------
x3 x1 x2 x4Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq - \sqrt{1 + \sqrt{2}}$$
$$x \geq - \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}} \wedge x \leq \sqrt{- \frac{1}{2} + \frac{\sqrt{5}}{2}}$$
$$x \geq \sqrt{1 + \sqrt{2}}$$
Solving inequality on a graph
Rapid solution
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/ / / 4 2 \ / 4 2 \ \ / / 4 2 \ \ / / 4 2 \ \\ Or\And\x <= CRootOf\x + x - 1, 1/, CRootOf\x + x - 1, 0/ <= x/, And\x <= CRootOf\x - 2*x - 1, 0/, -oo < x/, And\CRootOf\x - 2*x - 1, 1/ <= x, x < oo//
$$\left(x \leq \operatorname{CRootOf} {\left(x^{4} + x^{2} - 1, 1\right)} \wedge \operatorname{CRootOf} {\left(x^{4} + x^{2} - 1, 0\right)} \leq x\right) \vee \left(x \leq \operatorname{CRootOf} {\left(x^{4} - 2 x^{2} - 1, 0\right)} \wedge -\infty < x\right) \vee \left(\operatorname{CRootOf} {\left(x^{4} - 2 x^{2} - 1, 1\right)} \leq x \wedge x < \infty\right)$$
((-oo < x)∧(x <= CRootOf(x^4 - 2*x^2 - 1, 0)))∨((x < oo)∧(CRootOf(x^4 - 2*x^2 - 1, 1) <= x))∨((x <= CRootOf(x^4 + x^2 - 1, 1))∧(CRootOf(x^4 + x^2 - 1, 0) <= x))
Rapid solution 2
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/ 4 2 \ / 4 2 \ / 4 2 \ / 4 2 \ (-oo, CRootOf\x - 2*x - 1, 0/] U [CRootOf\x + x - 1, 0/, CRootOf\x + x - 1, 1/] U [CRootOf\x - 2*x - 1, 1/, oo)
$$x\ in\ \left(-\infty, \operatorname{CRootOf} {\left(x^{4} - 2 x^{2} - 1, 0\right)}\right] \cup \left[\operatorname{CRootOf} {\left(x^{4} + x^{2} - 1, 0\right)}, \operatorname{CRootOf} {\left(x^{4} + x^{2} - 1, 1\right)}\right] \cup \left[\operatorname{CRootOf} {\left(x^{4} - 2 x^{2} - 1, 1\right)}, \infty\right)$$
x in Union(Interval(-oo, CRootOf(x^4 - 2*x^2 - 1, 0)), Interval(CRootOf(x^4 - 2*x^2 - 1, 1), oo), Interval(CRootOf(x^4 + x^2 - 1, 0), CRootOf(x^4 + x^2 - 1, 1)))