Question

let a=x^2+4. rewrite the following equation in terms of a. (x^2+4)^2+32=12x^2+48 in resulting equation what is the coefficient of the a term in the resulting equation,what is the constant

Answer 1

We're given:
$a=x^2+4$

And with that we must rewrite the equation $(x^2+4)^2+32=12x^2+48$ in terms of $a$.

Currently, the equation is written in terms of $x^2$, so lets start by writing $x^2$ in terms of $a$:
$a=x^2+4$
$x^2=a-4$

Now lets replace every $x^2$ in $(x^2+4)^2+32=12x^2+48$ by $a-4$: 
$(x^2+4)^2+32=12x^2+48$
$(a-4+4)^2+32=12(a-4)+48$
$a^2+32=12(a-4)+48$

Now that we have the equation in terms of $a$, lets equal it to 0:
$a^2+32=12(a-4)+48$
$a^2+32-12(a-4)-48=0$
$a^2+32-12a+48-48=0$
$a^2-12a+32=0$

So the equation (in terms of $a$) is written in a standard form as $a^2-12a+32=0$.

We can clearly see that the coefficient of $a$ is -12 and the constant is 32.