Question

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Which statement is true about the solution of (picture below)?
A. x = –2 is an extraneous solution, and x = 6 is a true solution.
B. x = 6 is an extraneous solution, and x = –2 is a true solution.
C. Both x = –2 and x = 6 are extraneous solutions.
D. Both x = –2 and x = 6 are true solutions.

Answer 1

(D)

Explanation:

The given equation is:

`\sqrt[3]{x^2-12}=\sqrt[3]{4x}`

Cubing on both the sides, we get

`x^2-12=4x`

`x^2-4x-12=0`

`(x-6)(x+2)=0`

therefore, the solutions are:

`x=6` and
`x=-2`

Substitute x=6 in the given equation, we have

`(6)^2-12=4(6)`

`36-12=24`

`24=24`

which is true, thus x=6 is a true solution.

Put x=-2 in the given equation, we have

`(-2)^2-12=4(-2)`

`4-12=-8`

`-8=-8`

which is true, thus x=-2 is a true solution.

Therefore, Both x = –2 and x = 6 are true solutions.

Answer 2

`\sqrt[3]{x {}^(2) - 12 } = \sqrt[3]{4x}`
Cube by both sides:

`x {}^(2) - 12 = 4x`

`x {}^(2) - 4x - 12 = 0`

`(x - 6)(x + 2) = 0`
The solutions are:

`x = 6 \: \: and \: \: x = - 2`
The answer D is correct.