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Question Number 47331 by MJS last updated on 08/Nov/18

this remained unsolved \dots

∣ x - \frac{3}{4} ∣ \times ∣ x + \frac{5}{4} ∣= 3; x \in C

Commented by MJS last updated on 08/Nov/18

∣ a - \frac{3}{4} + b i ∣ \times ∣ a + \frac{5}{4} + b i ∣= 3 with a, b \in R

\sqrt{{(a - \frac{3}{4})}^{2} + b^{2}} \sqrt{{(a + \frac{5}{4})}^{2} + b^{2}} = 3

b^{4} + (2 a^{2} + a + \frac{17}{8}) b^{2} + (a^{4} + a^{3} - \frac{13}{8} a^{2} - \frac{15}{16} a - \frac{2079}{256}) = 0

b^{2} = - a^{2} - \frac{a}{2} - \frac{17}{16} \pm \sqrt{4 a^{2} + 2 a + \frac{37}{4}}

b \in R \Rightarrow b = \pm \sqrt{- a^{2} - \frac{a}{2} - \frac{17}{16} + \sqrt{4 a^{2} + 2 a + \frac{37}{4}}}

\sqrt{4 a^{2} + 2 a + \frac{37}{4}} ⩾ a^{2} + \frac{a}{2} + \frac{17}{16} \Rightarrow - \frac{9}{4} ⩽ a ⩽ \frac{7}{4}

solution is

x = a \pm i \sqrt{- a^{2} - \frac{a}{2} - \frac{17}{16} + \sqrt{4 a^{2} + 2 a + \frac{37}{4}}} with - \frac{9}{4} ⩽ a ⩽ \frac{7}{4}

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