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:
[tex]\sqrt{x-5}-\sqrt{x}=5-----(1)[/tex]
Now, we will solve this equation to obtain the solution as:
on squaring both side of the equation we obtain:
[tex](\sqrt{x-5}-\sqrt{x})^2=5^2[/tex]
[tex]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}[/tex]
Again on squaring both side of the equation we obtain:
[tex]x^2+225-30x=(x-5)x\\\\\x^2+225-30x=x^2-5x\\\\225=-5x+30x\\\\225=25x\\\\x=9[/tex]
Now when we plug x=9 back to the original equation i.e. equation (1) we get:
[tex]\sqrt{9-5}-\sqrt{9}=5\\\\\sqrt{4}-\sqrt{9}=5\\\\2-3=5\\\\-1=5[/tex]
Hence, the equation does not hold true.
Hence, 9 is a extraneous solution
Answer:
D.) x = 9 is an extraneous solution.
Step-by-step explanation:
