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Let a= x^2 +4. Rewrite the following equation in terms of a and set it equal to zero.
(x^2+4)^2+32=12x^2+48

which of the following are solutions for x?

A)-8
B)-2
C)4
D)0
E)2
F)-4
G)8

Respuesta :

MrsStrong

Answer:

C and G

Step-by-step explanation:

(x^2+4)^2+32=12x^2+48 is written as [tex](x^2+4)^2+32=12x^2+48[/tex].

We can simplify the expression using the term a by replacing x^2 + 4 with a.

[tex](x^2+4)^2+32=12x^2+48\\(x^2+4)^2+32=12(x^2+4)\\a^2 + 32 = 12a\\a^2 - 12a + 32 = 0[/tex]

This factors into (a-4)(a-8). Solve by setting each factor equal to 0 and solve for a.

a-4 = 0                 a=4

a-8=0                   a=8

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