Question-122360

Question Number 122360 by benjo_mathlover last updated on 16/Nov/20

Answered by $@y@m last updated on 16/Nov/20

P u t t = {(\sqrt{5 - 2 \sqrt{6}})}^{x}

t + \frac{1}{t} = 10

t^{2} - 10 t + 1 = 0

t = \frac{10 \pm \sqrt{100 - 4}}{2}

t = \frac{10 \pm 4 \sqrt{6}}{2} = 5 \pm 2 \sqrt{6}

\Rightarrow x = 2, - 2

Answered by nimnim last updated on 16/Nov/20

b y i n s p e c t i o n x = 2

Answered by liberty last updated on 16/Nov/20

consider : 5 - 2 \sqrt{6} = \frac{(5 - 2 \sqrt{6}) . (5 + 2 \sqrt{6})}{5 + 2 \sqrt{6}} = \frac{1}{5 + 2 \sqrt{6}}

or 5 - 2 \sqrt{6} = {(5 + 2 \sqrt{6})}^{- 1}

now let {(\sqrt{5 - 2 \sqrt{6}})}^{x} = p, gives

\Rightarrow p + p^{- 1} = 10 or \frac{p^{2} + 1}{p} = 10

\Rightarrow p^{2} - 10 p + 1 = 0; p_{1, 2} = \frac{10 \pm \sqrt{100 - 4}}{2}

\Rightarrow p_{1, 2} = 5 \pm 2 \sqrt{6}

for p_{1} = 5 + 2 \sqrt{6} = \sqrt{{(5 - 2 \sqrt{6})}^{x}}

\to 5 + 2 \sqrt{6} = \sqrt{{(5 + 2 \sqrt{6})}^{- x}}, we get x = - 2

similarly for p_{2} = 5 - 2 \sqrt{6} we get x = 2

thus the solutions is x = \pm 2 .

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