Question

Solve $\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98.$

• A. $\pm 1$
• B. $\pm 5$
• C. $\pm 4$
• D. $\pm 10$

Answer 1

The correct option is B $\pm 5$

$\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98$

$\Rightarrow$ $\frac{(x+\sqrt{x^2-1}{)}^2+(x-\sqrt{x^2-1}{)}^2}{(x{)}^2-(\sqrt{x^2-1)}{)}^2}=98$

$\Rightarrow$ $\frac{x^2+x^2-1+2x\sqrt{x^2-1}+x^2+x^2-1-2x\sqrt{x^2-1}}{(x{)}^2-(\sqrt{x^2-1}{)}^2}=98$

$\Rightarrow$ $\frac{4x^2-2}{x^2-x^2+1}=98$

$\Rightarrow$ $4x^2-2=98$

$\Rightarrow$ $4x^2=98+2$

$\Rightarrow$ $4x^2=100$

$\Rightarrow$ $x^2=25$

$\therefore$ $x=\pm 5$