I'll do a similar problem.
Solve for x: 6x^4 - 19x^2 + 15 = 0
Factor (using FOIL)
(2x^2 - 3)( 3x^2 - 5) = 0
If the product of two numbers is zero, at least one of them must be zero, so, set each equal to zero and solve for x.
2x^2 - 3 = 0
\(\displaystyle x^2 = \frac{3}{2}\)
Take square root of each side. (Don't forget the plus or minus sign.)
\(\displaystyle x = \pm\sqrt{\frac{3}{2}}\)
Rationalize the denominator.
\(\displaystyle x = \pm \sqrt{\frac{3\cdot 2}{2\cdot 2}}=\pm\frac{\sqrt 6}{2}\)
Then do the same for 3x^2 - 5 = 0
This gives you 4 roots. Be sure to check all four by plugging each back into the original equations and seeing if it tells the truth.
