Question

The roots of the equation ${(a+\sqrt{b})}^{x^2-15}+{(a-\sqrt{b})}^{x^2-15}=2a,$ where $a^2-b=1$ are

• A. $\pm 2,\pm \sqrt{3}$
• B. $\pm 4,\pm \sqrt{14}$
• C. $\pm 3,\pm \sqrt{5}$
• D. $\pm 6,\pm \sqrt{20}$

Answer 1

The correct option is B $\pm 4,\pm \sqrt{14}$

Given equation, ${(a+\sqrt{b})}^{x^2-15}+{(a-\sqrt{b})}^{x^2-15}=2a$ ....(1)
Since, $a^2-b=1$
$\Rightarrow (a+\sqrt{b})(a-\sqrt{b})=1$
$a+\sqrt{b}=\frac{1}{a-\sqrt{b}}$
Using this in equation (1),
${\left(\frac{1}{a-\sqrt{b}}\right)}^{x^2-15}+{(a-\sqrt{b})}^{x^2-15}=2a$
$\Rightarrow {(a-\sqrt{b})}^{2(x^2-15)}-2a{(a-\sqrt{b})}^{x^2-15}+1=0$
Put ${(a-\sqrt{b})}^{x^2-15}=t$
$\Rightarrow t^2-2at+1=0$
$\Rightarrow t=\frac{2a\pm \sqrt{4(a^2-1)}}{2}$
$\Rightarrow t=\frac{2a\pm 2\sqrt{b}}{2}$
$t=a\pm \sqrt{b}$
$\Rightarrow {(a-\sqrt{b})}^{x^2-15}=(a-\sqrt{b})or\frac{1}{a-\sqrt{b}}$
$\Rightarrow x^2-15=\pm 1$
$\Rightarrow x^2=16,14$
$\Rightarrow x=\pm 4,\pm \sqrt{14}$
Hence, option 'B' is correct.