Mathematics: 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.

Consider the radical equation square root of n+4 =

RealGibbon857

10.03.2022

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.

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GreenHerbz206

29.05.2020

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

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hellokitty1647

25.06.2021

is it 17/25 ? give thanks if it's crt ✔️ if not then mark report .

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