Question

If ${\left(5-2\sqrt{6}\right)}^{x^{2-1}}+{\left(5+2\sqrt{6}\right)}^{x^2-1}=10,$ when the value of x are

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. 0
• D. all of these above

Answer 1

Answer: (D)

all of these above

Solution:-

Given $(5-2\sqrt{6}{)}^{x^2-1}+(5+2\sqrt{6}{)}^{x^2-1}=10$

we have $5-2\sqrt{6}=\frac{1}{5+2\sqrt{6}}$

if $5-2\sqrt{6}=a$and $x^2-1=y$

then $a^y+\frac{1}{a^y}=10$

Multiplying $a^y$ throughout, we get

$(a^y{)}^2-10a^y+1=0$

Let $a^y=z$

then $z^2-10z+1=0$

$\Rightarrow z=\frac{-(-10)\pm \sqrt{{10}^2-4}}{2\times 1}=5\pm 2\sqrt{6}$

$\Rightarrow z=a^y=5\pm 2\sqrt{6}$

so $(5-2\sqrt{6}{)}^y=(5-2\sqrt{6})\Rightarrow y=1$

and $(5-2\sqrt{6})=(5+2\sqrt{6})=y=-1$

$\Rightarrow x^2-1=\pm 1$

$\Rightarrow x=0,\pm \sqrt{2}$