contestada

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

Respuesta :

We're given:
[tex]a=x^2+4[/tex]

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

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

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

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

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

We can clearly see that the coefficient of [tex]a[/tex] is -12 and the constant is 32.