You can put this solution on YOUR website! How can we put this in familiar form?
9 and 4 are perfect squares. These are hints at something.
we know a^2-b^2 = (a+b)(a-b)
we can do this! x^2^2-3^2=(x^2+3)(x^2-3)=0
3^2+3=0, x=Square root of 3i
3^2-3=0, x=Square root of 3
This is not as complicated as it looks, but you must use the common products of Algebra AND be on the look-out for perfect squares either in the exponents or the real numbers themselves.
square root of 3i AND the square root of 3 are the answers.
You can put this solution on YOUR website! To be solved there must be an equal sign.
Otherwise it can be factored or simplified but not solved.
let u^2=x^4 so u=x^2
u^2-9=
(u-3)*(u+3)
(x^2-3)*(x^2+3)
(
