# Thread: Solving a quartic equation

1. ## Solving a quartic equation

This problem seems a bit out of the ordinary. s^4+ s^2-6 =0

2. Originally Posted by sgonzalez90
This problem seems a bit out of the ordinary. s^4+ s^2 -6 =0
This is a quadratic in $\displaystyle s^2$. Let $\displaystyle w = s^2$:

$\displaystyle w^2 + w - 6 = 0$.

Solve for $\displaystyle w$. Hence solve for $\displaystyle s$.

3. Oh my gosh I'm so sorry, i typed it wrong again. I'm sorry im really tired and had a long day. I'm just trying to brush up on my algebra before i begin my calculus course in two weeks.

s^4+ 3s^2 -6 =0
this is the correct problem i promise, thank you.

4. Originally Posted by sgonzalez90
Oh my gosh I'm so sorry, i typed it wrong again. I'm sorry im really tired and had a long day. I'm just trying to brush up on my algebra before i begin my calculus course in two weeks.

s^4+ 3s^2 -6 =0
this is the correct problem i promise, thank you.
Just change the appropriate numbers in post #2.

5. Originally Posted by sgonzalez90
s^4+ 3s^2 -6 =0
this is the correct problem i promise, thank you.
This one won't factor, so apply the Quadratic Formula:

$\displaystyle x^2\, =\, \frac{-(3)\, \pm\, \sqrt{(3)^2\, -\, 4(1)(-6)}}{2(1)}$

$\displaystyle x^2\, =\, \frac{-3\, \pm\, \sqrt{9\, +\, 24}}{2}$

$\displaystyle x^2\, =\, \frac{-3\, \pm\, \sqrt{33}}{2}$

Then take the square root of either side to solve for "x=". Yes, you'll have some nested radicals.