Question

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

A. The solution n = 5 is an extraneous solution.
B.Both n = 5 and n = 0 are true solutions.
C.The solution n = 0 is an extraneous solution.
D. Neither are true solutions to the equation.

Answer 1

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

Answer 2

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

Explanation:

The given equation is
`√(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:

`√(0+4) =0-2` False .

For n=5.

`√(5+4) =5-2`

True.

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