How can such questions be solved.? x^2 −∣7∣ +10=0 x^2 −∣x∣−6>0


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Question Number 64015 by Rio Michael last updated on 12/Jul/19

$H o w c a n s u c h q u e s t i o n s b e s o l v e d . ?$ $x^{2} - ∣ 7 ∣ + 10 = 0$ $x^{2} - ∣ x ∣ - 6 > 0$
	Answered by MJS last updated on 12/Jul/19

	$x^{2} - ∣ x ∣ - 6 = 0$ $case 1 : x < 0 \Rightarrow ∣ x ∣= - x$ $x^{2} + x - 6 = 0 \Rightarrow x = - 3 \lor x = 2 but x < 0 \Rightarrow x = - 3$ $case 2 : x ⩾ 0 \Rightarrow ∣ x ∣= x$ $x^{2} - x - 6 = 0 \Rightarrow x = - 2 \lor x = 3 but x ⩾ 0 \Rightarrow x = 3$ $x^{2} - ∣ x ∣ - 6 = 0 \Rightarrow x = \pm 3$ $x^{2} - ∣ x ∣ - 6 > 0 \Rightarrow x < - 3 \lor x > 3$
	Commented by Rio Michael last updated on 12/Jul/19

	$d o e s i t a l s o a p p l y t o x^{2} + ∣ 7 x ∣ + 10 = 0 ?$
	Commented by MJS last updated on 12/Jul/19

	$x^{2} + ∣ 7 x ∣ + 10 = 0$ $x^{2} + 7 ∣ x ∣ + 10 = 0$ $case 1 x < 0 \Rightarrow ∣ x ∣= - x$ $x^{2} - 7 x + 10 = 0 \Rightarrow x = 2 \lor x = 5 but x < 0 \Rightarrow no solution$ $case 2 x ⩾ 0 \Rightarrow ∣ x ∣= x$ $x^{2} + 7 x + 10 = 0 \Rightarrow x = - 2 \lor x = - 5 but x ⩾ 0 \Rightarrow no solution$ $the absolute minimum of x^{2} + 7 ∣ x ∣ + 10 is 10$
	Commented by MJS last updated on 12/Jul/19

	$we might get complex solutions$ $x = a + b i$ $leads to$ ${(a + b i)}^{2} + 7 ∣ a + b i ∣ + 10 = 0$ $a^{2} - b^{2} + 7 \sqrt{a^{2} + b^{2}} + 10 + 2 a b i = 0$ $\Rightarrow a^{2} - b^{2} + 7 \sqrt{a^{2} + b^{2}} + 10 = 0 \land 2 a b i = 0$ $2 a b i = 0 \Rightarrow a = 0 \lor b = 0$ $a = 0$ $- b^{2} + 7 ∣ b ∣ + 10 = 0 \Rightarrow again 2 cases \Rightarrow$ $\Rightarrow b = - \frac{7}{2} - \frac{\sqrt{89}}{2} \lor b = \frac{7}{2} + \frac{\sqrt{89}}{2}$ $\Rightarrow x = \pm (\frac{7}{2} + \frac{\sqrt{89}}{2}) i$ $b = 0 leads to the above equation only with$ $a instead of x \Rightarrow no real solutions$
	Commented by Rio Michael last updated on 12/Jul/19

	$t h a n k y o u$
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