Question

Find all real solutions for the following equation.
x^4-x^2=3x^2+45

Answer 1

x^4-x^2=3x^2+45
add -3x^2:
x^4 - 4x^2 = 45
add -45:
x^4 - 4x^2 - 45 = 0
x^4 - 9x^2 + 5x^2 - 45 = 0
x^2(x^2 - 9) + 5(x^2 - 9) = 0
(x^2-9)(x^2+5) = 0
(x-3)(x+3)(x^2+5) = 0
can you solve it now?

Answer 2

Ok thanks. But where did the 9 come from???? For my solution I got x=-3,3 and i to the square root of 5.

Answer 3

\[x = -3,~~~x = 3, ~~~ x=-\sqrt{5}i, ~~~ x = \sqrt{5}i\]But here they want only the real solutions and so the answer is is just \(x=-3\) and \(x = 3\).

Do answer your question where did the 9 come from?
In factoring the left hand side of x^4 - 4x^2 - 45 = 0 I used the AC method
a = 1, b = -4, c = -45
a * c = 1 * (-45) = -45
Split -45 into two factors such that the product is -45 and the sum is the middle coefficient -4. -9 and 5 add to -4 and multiply to -45:
x^4 - 4x^2 - 45 = 0
x^4 - 9x^2 + 5x^2 - 45 = 0
....
as before

Answer 4

Thank you!