Question

lf $(7+4\sqrt{3}{)}^{x^2-8}+(7-4\sqrt{3}{)}^{x^2-8}=14$, then $x$ is

• A. $3,\sqrt{7}$
• B. $\pm 3,\pm 1$
• C. $\pm 3;\pm \sqrt{7}$
• D. $\pm 3,\pm 4$

Answer 1

The correct option is C $\pm 3;\pm \sqrt{7}$

We have,
$(7+4\sqrt{3}{)}^{x^2-8}+(7-4\sqrt{3}{)}^{x^2-8}=14$
Also $(7+4\sqrt{3})(7-4\sqrt{3})=7^2-16\times 3=1$
Thus above equation becomes,
$(7+4\sqrt{3}{)}^{x^2-8}+{\left(\frac{1}{7+4\sqrt{3}}\right)}^{x^2-8}=14$
Substitute $(7+4\sqrt{3}{)}^{x^2-8}=y$
$\Rightarrow y^2-14y+1=0$
$\Rightarrow y=\frac{14\pm \sqrt{{14}^2-4}}{2}=7\pm 4\sqrt{3}=(7+4\sqrt{3}{)}^{\pm 1}$
$\therefore (7+4\sqrt{3}{)}^{x^2-8}=(7+4\sqrt{3}{)}^{\pm 1}$
$\Rightarrow x^2-8=\pm 1\Rightarrow x^2=9,7$
$\Rightarrow x=\pm 3,\pm \sqrt{7}$