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  4. Find the number of real solutions of (7+4 √3)|x|-8+(7-4 √3)|x|-8=14.

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Q. Find the number of real solutions of $(7+4 \sqrt{3})^{|x|-8}+(7-4 \sqrt{3})^{|x|-8}=14$.

Complex Numbers and Quadratic Equations

Solution:

$(7+4 \sqrt{3})^{|x|-8}+(7-4 \sqrt{3})^{|x|-8}=14$
$\Rightarrow(7+4 \sqrt{3})^{|x|-8}+\frac{1}{(7+4 \sqrt{3})^{|x|-8}}=14$
$\ldots\left[\Rightarrow 7-4 \sqrt{3}=\frac{1}{7+4 \sqrt{3}}\right]$
$(7+4 \sqrt{3})^{|x|-8}=t$, (say)
$\Rightarrow t+\frac{1}{t}=14$
$\Rightarrow t^{2}-14 t+1=0$
$\Rightarrow t=\frac{14 \pm \sqrt{196-4}}{2}$
$=\frac{14 \pm \sqrt{192}}{2}$
$=7 \pm 4 \sqrt{3}$
$\Rightarrow t=7+4 \sqrt{3}$
or $t=7-4 \sqrt{3}=\frac{1}{7+4 \sqrt{3}}$
$\Rightarrow(7+4 \sqrt{3})^{|x|-8}=(7+4 \sqrt{3})^{1}$
or $(7+4 \sqrt{3})^{|x|-8}=(7+4 \sqrt{3})^{-1}$
$\Rightarrow|x|-8=1 $
or $ |x|-8=-1 $
$ \Rightarrow|x|=9 $
or $ |x|=7 $
$ \Rightarrow x=9,-9 $
or $x=7,-7$
$\Rightarrow$ Number of real solutions $=4$