Absolute Power

Algebra Level 4

$\large (7+4\sqrt{3})^{|x|-8}+(7-4\sqrt{3})^{|x|-8}=14$ The number of real roots of the above equation is:

The answer is 4.

1 solution

Rohit Udaiwal
Oct 30, 2015

$7-4\sqrt{3}=\dfrac{1}{7+4\sqrt{3}}$

If $a=(7+4\sqrt{3})^{|x|-8}$ then $a+\dfrac{1}{a}=14$

$\implies a^2-14a+1=0 \\ \implies a=\dfrac{14 \pm \sqrt{196-4}}{2}=\dfrac{14 \pm 8\sqrt{3}}{2}=7 \pm 4\sqrt{3} \\ \color{#20A900}{\therefore } (7+4\sqrt{3})^{|x|-8}=(7+4\sqrt{3})^{\pm 1}\\ \therefore |x|-8=\pm 1 \implies |x|=9,7 \\ \therefore x=\pm 9,\pm 7 \\ \therefore 4 ~ \text{roots.}$

overrated problem

Rishi Sharma - 5 years, 7 months ago

14^2 in 196 not 194 rest the solution is nice:)

Shreyansh Choudhary - 5 years, 7 months ago

Absolute Power

The answer is 4.

1 solution

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