Ask Me Help Desk - Solve a "quadratic type" equation

Quote:

Originally Posted by brosh4

3y4 = 28y2 - 9

I'm assuming you mean this:

$<br /> 3 y^4 = 28y^2 - 9<br />$

Start by substituting a new variable for y^2. If we let w = y^2, then this becomes

$<br /> 3w^2 = 28y -9<br />$

You can rearrange to get this into a standard quadratic form, then use the quadratic formula to solve for w. There will be two soultions for w. Then finally y = sqrt(w), but be careful to check whether both positive and negative values of sqrt(w) work in the original equation.