Question

Consider the radical equation square root of n+4 = n – 2. Which statement is true about the solutions n = 5 and n = 0?

Answer 1

Answer:

C.The solution n = 0 is an extraneous solution.

Step-by-step explanation:

The given equation is $\sqrt{n+4} =n-2.$

Squaring both sides we have:

$n+4=(n-2)^{2}$

Expanding the right side:

$n+4=n^{2} -4n+4$

Or ,

$n^{2} -5n=0$

n(n-5)=0

n=0 or n=5.

Checking the solution for n=0 in the above equation:

$\sqrt{0+4} =0-2$ False .

For n=5.

$\sqrt{5+4} =5-2$

True.

C.The solution n = 0 is an extraneous solution.

Answer 2

it's not D i took the test and got it wrong