Question

Let a=x^2+4. Use a to find the solutions for the following equation: (x^2+4)^2+32=12x^2+48. Which one of the following are solutions for x? Select any/all that apply. -8, -2, 4, 0, 2, -4, 8

Answer 1

-2,0,2

Explanation:

The given equation is:

`(x^2+4)^(2)+32=12x^2+48`

⇒
`(x^2+4)^(2)+32=12(x^2+4)`

Substituting
`(x^2+4)=a` in the above equation, we get

⇒
`a^(2)+32=12a`

⇒
`a^2-12a+32=0`

⇒
`a^2-4a-8a+32=0`

⇒
`a(a-4)-8(a-4)=0`

⇒
`(a-8)(a-4)=0`

⇒
`a=8,4`

Now,
`(x^2+4)=a`, then substituting the value of a in this equation,

`x^(2)+4=8` and
`x^2+4=4`

⇒
`x^(2)+4=8`

⇒
`x={\pm}2` and

⇒
`x^(2)+4=4`

⇒
`x=0`

Thus, the value of x are -2,0 and 2.

Answer 2

`(x^2+4)^2+32=12x^2+48 \\ (x^2+4)^2+32=12(x^2+4) \ \ \ |-12(x^2+4) \\ (x^2+4)^2-12(x^2+4)+32=0 \\ \hbox{substitute a for } x^2+4: \\ a^2-12a+32=0 \\ a^2-4a-8a+32=0 \\ a(a-4)-8(a-4)=0 \\ (a-8)(a-4)=0 \\ a-8=0 \ \lor \ a-4=0 \\ a=8 \ \lor \ a=4 \\ \\ \hbox{substitute 8 and 4 for a and solve for x:} \\ a=8 \\ \Downarrow \\ 8=x^2+4 \ \ \ |-4 \\ 4=x^2 \\ x=-2 \ \lor \ x=2 \\ \\ a=4 \\ \Downarrow \\ 4=x^2+4 \ \ \ |-4 \\ 0=x^2 \\ x=0 \\ \\ \boxed{x=-2 \hbox{ or } x=0 \hbox{ or } x=2}`

The solutions for x are -2, 0, 2.