Answer :
[tex](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}[/tex]
The solutions for x are -2, 0, 2.
The solutions for x are -2, 0, 2.
Answer:
-2,0,2
Step-by-step explanation:
The given equation is:
[tex](x^2+4)^{2}+32=12x^2+48[/tex]
⇒[tex](x^2+4)^{2}+32=12(x^2+4)[/tex]
Substituting [tex](x^2+4)=a[/tex] in the above equation, we get
⇒[tex]a^{2}+32=12a[/tex]
⇒[tex]a^2-12a+32=0[/tex]
⇒[tex]a^2-4a-8a+32=0[/tex]
⇒[tex]a(a-4)-8(a-4)=0[/tex]
⇒[tex](a-8)(a-4)=0[/tex]
⇒[tex]a=8,4[/tex]
Now, [tex](x^2+4)=a[/tex], then substituting the value of a in this equation,
[tex]x^{2}+4=8[/tex] and [tex]x^2+4=4[/tex]
⇒[tex]x^{2}+4=8[/tex]
⇒[tex]x={\pm}2[/tex] and
⇒[tex]x^{2}+4=4[/tex]
⇒[tex]x=0[/tex]
Thus, the value of x are -2,0 and 2.