Question

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

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

Answer 1

The correct option is B

$\pm 4,\pm \sqrt{14}$

Explanation for the correct option:

Find the roots of the given equation

Let $y=a+\sqrt{b}$

$$\begin{gathered} \frac{1}{y}=\frac{1}{a+\sqrt{b}} \\ =\frac{a-\sqrt{b}}{\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right)} \\ =\frac{a-\sqrt{b}}{a^2-b} \\ =a-\sqrt{b}\left[\because a^2-b=1\right] \end{gathered}$$

$\therefore {\left(y\right)}^{x^2-15}+{\left(\frac{1}{y}\right)}^{x^2-15}=2a$

Let ${\left(y\right)}^{x^2-15}=t$

$$\begin{gathered} \therefore t+\frac{1}{t}=2a \\ \Rightarrow t^2-2at+1=0 \\ t=\frac{2a\pm \sqrt{4a^2-4}}{2} \\ =a\pm \sqrt{a^2-1} \\ =a\pm \sqrt{b}\left[\because a^2-b=1\right] \end{gathered}$$

$t=a+\sqrt{b}$ [by taking positive sign]

$$\begin{gathered} \Rightarrow {\left(y\right)}^{x^2-15}=a+\sqrt{b}\left[\because {\left(y\right)}^{x^2-15}=t\right] \\ \Rightarrow {\left(a+\sqrt{b}\right)}^{x^2-15}=a+\sqrt{b}\left[\because a+\sqrt{b}=y\right] \\ \Rightarrow x^2-15=1 \\ \Rightarrow x^2=16 \\ \Rightarrow x=\pm 4 \end{gathered}$$

$t=a-\sqrt{b}$ [by taking negative sign]

$$\begin{gathered} \Rightarrow {\left(y\right)}^{x^2-15}=a-\sqrt{b}\left[\because {\left(y\right)}^{x^2-15}=t\right] \\ \Rightarrow {\left(a-\sqrt{b}\right)}^{x^2-15}={\left(a-\sqrt{b}\right)}^{-1}\left[\because a-\sqrt{b}=\left(\frac{1}{y}\right)\right] \\ \Rightarrow x^2-15=-1 \\ \Rightarrow x^2=14 \\ \Rightarrow x=\pm \sqrt{14} \end{gathered}$$

Therefore $x=\pm 4,\pm \sqrt{14}$

Hence, the correct option is $\left(\mathrm{B}\right)$.