Question

Please help!!! will mark brainliest

Answer 1

A=x^2+4
(x^2+4)^2+32=12x^2+48
12x^2+48=12(x^2+4)
(x^2+4)^2+32=12(x^2+4)
a^2+32=12a
subtract 12a from both sides
a^2-12a+32=0
factor
(a-4)(a-8)=0
set each to zero
a-4=0
a=4
a-8=0
a=8
a=4 or 8
a=x^2+4

8=x^2+4 and
4=x^2+4
subtract 4
4=x^2
0=x^2
square root
+/-2=x
0=x
x=-2,0,2
ax^2+bx+c=0
coefficient of the a term is 1
constant=32

Answer 2

A = x² + 4
& (x² + 4) +32 = 12x² + 48. You can factorize the left side of this equation

So 12x² + 48 = 12(x² + 4)==> (x² + 4) +32 = 12(x² + 4)
Now replace (x² +4) by a ==> a +32 = 12a
transpose all on the left side: a+32 -12a = 0==> -11a + 32 = 0
If you are asked to calculate then -11a=-32 & a=32/11