Question

NEED ANSWER NOW PLEASEEEEEEEEEEEEE

Answer 1

Answer:

(D)

Step-by-step 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.