Question

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 what are the solutions for x

Answer 1

Just replace x^2+4 with a
factor out on left side
(x^2+4)^2+32=12(x^2+4)
replace x^2+4 with a
a^2+32=12a
minus 12a both sides
a^2-12a+32=0
solve
factor
what 2 numbers mulitply to get 32 and add to get -12
-4 and -8
(a-4)(a-8)=0

set each to zero
a-4=0
a=4

a-8=0
a=8


remember
a=x^2+4
so

4=x^2+4
0=x^2
0=x

8=x^2+4
4=x^2
+/-2=x


x=-2, 0, 2