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NEED ANSWER NOW PLEASEEEEEEEEEEEEE
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.

NEED ANSWER NOW PLEASEEEEEEEEEEEEE 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 class=

Respuesta :


[tex] \sqrt[3]{x {}^{2} - 12 } = \sqrt[3]{4x} [/tex]
Cube by both sides:
[tex]x {}^{2} - 12 = 4x[/tex]
[tex]x {}^{2} - 4x - 12 = 0[/tex]
[tex](x - 6)(x + 2) = 0[/tex]
The solutions are:
[tex]x = 6 \: \: and \: \: x = - 2[/tex]
The answer D is correct.
boffeemadrid

Answer:

(D)

Step-by-step explanation:

The given equation is:

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

Cubing on both the sides, we get

[tex]x^2-12=4x[/tex]

[tex]x^2-4x-12=0[/tex]

[tex](x-6)(x+2)=0[/tex]

therefore, the solutions are:

[tex]x=6[/tex] and [tex]x=-2[/tex]

Substitute x=6 in the given equation, we have

[tex](6)^2-12=4(6)[/tex]

[tex]36-12=24[/tex]

[tex]24=24[/tex]

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

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

[tex](-2)^2-12=4(-2)[/tex]

[tex]4-12=-8[/tex]

[tex]-8=-8[/tex]

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

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

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