solve for x∈C: ∣x−(3/4)∣×∣x+(5/4)∣=3


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Question Number 47291 by MJS last updated on 07/Nov/18

$solve for x \in C :$ $∣ x - \frac{3}{4} ∣ \times ∣ x + \frac{5}{4} ∣= 3$
	Answered by MrW3 last updated on 08/Nov/18

	$l e t x + \frac{1}{4} = a + b i$ $∣ x - \frac{3}{4} ∣ \times ∣ x + \frac{5}{4} ∣= 3$ $\sqrt{{(a - 1)}^{2} + b^{2}} \times \sqrt{{(a + 1)}^{2} + b^{2}} = 3$ $[{(a - 1)}^{2} + b^{2}] \times [{(a + 1)}^{2} + b^{2}] = 9$ ${(a^{2} - 1)}^{2} + b^{2} [{(a - 1)}^{2} + {(a + 1)}^{2}] + b^{4} = 9$ $b^{4} + 2 (a^{2} + 1) b^{2} + (a^{2} + 2) (a^{2} - 4) = 0$ $b^{2} = - (a^{2} + 1) \pm \sqrt{{(a^{2} + 1)}^{2} - (a^{2} + 2) (a^{2} - 4)}$ $b^{2} = - (a^{2} + 1) \pm \sqrt{4 a^{2} + 9}$ $b^{2} ⩾ 0$ $- (a^{2} + 1) + \sqrt{4 a^{2} + 9} ⩾ 0$ $\sqrt{4 a^{2} + 9} ⩾ a^{2} + 1$ $4 a^{2} + 9 ⩾ a^{4} + 2 a^{2} + 1$ $a^{4} - 2 a^{2} - 8 ⩽ 0$ $(a^{2} + 2) (a^{2} - 4) ⩽ 0$ $a^{2} - 4 ⩽ 0$ $\Rightarrow - 2 ⩽ a ⩽ 2$ $\Rightarrow b^{2} = \sqrt{4 a^{2} + 9} - a^{2} - 1$ $\Rightarrow b = \pm \sqrt{\sqrt{4 a^{2} + 9} - a^{2} - 1}$ $\Rightarrow x = (a - \frac{1}{4}) \pm i \sqrt{\sqrt{4 a^{2} + 9} - a^{2} - 1}$ $w i t h - 2 ⩽ a ⩽ 2$
	Commented by MJS last updated on 09/Nov/18

	$thank you!$
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