Question

If ${(5+2\sqrt{6})}^{x^2-3}+{(5-2\sqrt{6})}^{x^2-3}=10$, then $x=$.........

• A. $2,2$
• B. $\sqrt{2},-\sqrt{2}$
• C. $\sqrt{2},+\sqrt{2}$
• D. $2,-2$

Answer 1

The correct option is D $2,-2$

Given,

$(5+2\sqrt{6}{)}^{x^2-3}+(5-2\sqrt{6}{)}^{x^2-3}=10$

Taking exponential form on both sides, we get,

$e^{(5+2\sqrt{6}{)}^{x^2-3}+(5-2\sqrt{6}{)}^{x^2-3}}=e^{10}$

$e^{(5+2\sqrt{6}{)}^{x^2-3}}\cdot e^{(5-2\sqrt{6}{)}^{x^2-3}}=e^{10}$

$[e^{(5+2\sqrt{6})}\cdot e^{(5-2\sqrt{6})}{]}^{x^2-3}=e^{10}$

$[e^{10}{]}^{x^2-3}=[e^{10}{]}^1$

$x^2-3=1$

$x^2=4$

$x=\pm 2$