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.