solve-for-x-C-x-3-4-x-5-4-3-

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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