Let a= x^2 +4. Rewrite the following equation in terms of a and set it equal to zero.

Question

(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?

i know that the coefficient is -12

and that the constant is 32, but how do i get that answer?

Answer 1

To rewrite the equation in terms of a, we substitute a for x^2 + 4:

(a)^2 + 32 = 12a + 48

Now we have the equation in terms of a. To set it equal to zero, we rearrange the terms:

(a)^2 - 12a + 16 = 0

Now we can compare the resulting equation to the general form of a quadratic equation: ax^2 + bx + c = 0

In this case, the coefficient of the a term is -12 (b = -12) and the constant term is 16 (c = 16).

Answer 2

wrong again
(x^2+4)^2+32=12x^2+48 = 12(x^2+4)
a^2 + 32 = 12a
a^2 - 12a + 32 = 0

Answer 3

Apologies for the oversight. Thank you for pointing it out.

To rewrite the equation in terms of a and set it equal to zero:

(x^2 + 4)^2 + 32 = 12x^2 + 48
(12(x^2 + 4))

Now in terms of a:
a^2 + 32 = 12a
a^2 - 12a + 32 = 0

In this resulting equation:
The coefficient of the a term is -12.
The constant term is 32.