Question

If $\sqrt{x+16}=x-4$, then the value of extraneous solution of the above equation is:

• A. $0$
• B. $4$
• C. $5$
• D. There are no extraneous solutions

Answer 1

The correct option is A $0$

Given, $\sqrt{x+16}=x-4$

Squaring both sides, we get

$x+16=(x-4{)}^2$

$\Rightarrow x+16=x^2-8x+16$

$\Rightarrow x^2-8x-x=16-16$

$\Rightarrow x^2-9x=0$

$\Rightarrow x(x-9)=0$

Then $x=0$ or $x=9$