Question

Solve the following equation $(x\ge 0)$

$\frac{\sqrt{x+13}+\sqrt{x-1}}{\sqrt{x+13}-\sqrt{x-1}}=3$
solution is $\left\{\frac{17}{3}\right\}$
If true then enter $1$ and if false then enter $0$

Answer 1

$\frac{\sqrt{x+13}+\sqrt{x-1}}{\sqrt{x+13}-\sqrt{x-1}}=3=\frac{\sqrt{x+13}+\sqrt{x-1}}{\sqrt{x+13}-\sqrt{x-1}}\ast \frac{\sqrt{x+13}+\sqrt{x-1}}{\sqrt{x+13}+\sqrt{x-1}}=\frac{x+13+x-1+2\sqrt{x+13}\ast \sqrt{x-1}}{x+13-x+1}=\frac{x+6+\sqrt{x+13}\sqrt{x-1}}{7}=3=\sqrt{x+13}\sqrt{x-1}=15-xSquaring,=(x+13)(x-1)=(15-x)(15-x)=x^2-13+12x=x^2-30x+225=42x=238\therefore x=\frac{17}{3}$