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

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

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

Answer 2

Answer:

-2,0,2

Step-by-step 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.