Let a = x^2 plus 4. use a to find the solutions for the following equation: (x^2 plus. 4)^2 plus 32 = 12x^2 plus 48 which of the following are solutions for x

0 or +-2

Step-by-step explanation:

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

=> x^4 + 8x^2 + 16 + 32 - 12x^2 - 48 = 0

=> x^4 - 4x^2 = 0

=> x^2(x^2 - 4) = 0

=> x^2 = 0 or 4

=> x = 0 or +-2

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

c.

step-by-step explanation: