Question

Which statement about square root x-5 - square root x = 5 is true?
a) x = –3 is a true solution.
b) x = –3 is an extraneous solution.
c) x = 9 is a true solution.
d) x = 9 is an extraneous solution.

Answer 1

Hello,

Answer D

$\sqrt{x-5} - \sqrt{x} =5$
==>$(x-5) -2*\sqrt{(x-5)x}+x =25$
==>$2x-30=\sqrt{x(x-5)}$
==>$x-15=\sqrt{x(x-5)}$
==>$x^2-30x+225=x(x-5)$
==>$225=-5x+30x$
==>$25x=225$
==>$x=9$
But:

$\sqrt{9-5} - \sqrt{9}=\sqrt{4} - \sqrt{9}=2-3=-1$≠5

Explanation:

$\lim_{x \to \infty} \sqrt{x-5}- \sqrt{x}$

$= \lim_{x \to \infty} \frac{(\sqrt{x-5}- \sqrt{x})*(\sqrt{x-5}+ \sqrt{x})} {\sqrt{x-5}+ \sqrt{x}}$

$= \lim_{x \to \infty} \frac{(-5)}{\sqrt{x-5}+ \sqrt{x}}$

$= \lim_{x \to \infty} \frac{-5}{\infty}=0$