Question

Which statement about sqrt x-5 minus sqrt x=5 true?

Answer 1

Answer:

9 is an extraneous solution.

Step-by-step explanation:

Extraneous solution is a solution that when plugged into the equation do not hold true.

Here we are given an algebraic equation in terms of single variable x as:

$\sqrt{x-5}-\sqrt{x}=5-----(1)$

Now, we will solve this equation to obtain the solution as:

on squaring both side of the equation we obtain:

$(\sqrt{x-5}-\sqrt{x})^2=5^2$

$x-5+x-2\sqrt{x-5}\sqrt{x}=25\\\\2x-5-2\sqrt{x-5}\sqrt{x}=25\\\\2x-30=2\sqrt{x-5}\sqrt{x}\\\\on\ dividing\ both\ side\ by\ 2\ we\ obtain:\\\\x-15=\sqrt{x-5}\sqrt{x}$

Again on squaring both side of the equation we obtain:

$x^2+225-30x=(x-5)x\\\\\x^2+225-30x=x^2-5x\\\\225=-5x+30x\\\\225=25x\\\\x=9$

Now when we plug x=9 back to the original equation i.e. equation (1) we get:

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

Hence, the equation does not hold true.

Hence, 9 is a extraneous solution