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Question Number 96311 by bemath last updated on 31/May/20

$${(4+\sqrt{15})}^x+{(4-\sqrt{15})}^x=62$$$$x=?$$

Answered by bobhans last updated on 31/May/20

$$4-\sqrt{15}\times \frac{4+\sqrt{15}}{4+\sqrt{15}}=\frac{1}{4+\sqrt{15}}$$$$\mathrm{so}{(4+\sqrt{15})}^x+\frac{1}{{(4+\sqrt{15})}^x}=62$$$$t+\frac{1}{t}-62=0,[\mathrm{t}={(4+\sqrt{15})}^x]$$$$t^2-62t+1=0\Rightarrow \mathrm{t}=\frac{62+16\sqrt{5}}{2}=31+8\sqrt{5}$$$$={(4+\sqrt{15})}^2={(4+\sqrt{15})}^x,\mathrm{we}\mathrm{get}x=2$$$$theotherrootis{(4+\sqrt{15})}^{-2}={(4+\sqrt{15})}^x$$$$x=-2$$$$$$

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