Question

Working through this equation.

Answer 1

\[3y^4-9y^2+1=0\]

Answer 2

How would I go about solving this. I was thinking I should move the 1 over and take a 3y^2 out of what is left then go from there?

Answer 3

let x= y^2

then your equation is
3x^2 -9x +1= 0

this is a quadratic,
http://www.purplemath.com/modules/solvquad4.htm
you can use the quadratic formula to solve for x
then y = ± sqr(x)

Answer 4

I get

\[\frac{ 9 +/- \sqrt{69} }{ 6 }\]

?

Answer 5

yes those are the two roots of x

y= +sqr(x) and -sqr(x)
you will get 4 different values for y

Answer 6

still kinda lost, I square the whole x value I got now?

Answer 7

you solved
3x^2 -9x +1= 0

and got two values for x

remember we let y^2 = x
so y = ±sqr(x)

Answer 8

in other words, one of the roots is
\[ y = \sqrt{ \frac{1}{6}\left(9- \sqrt{69}\right) }\]
another is
\[ y = -\sqrt{ \frac{1}{6}\left(9- \sqrt{69}\right) }\]

the other two roots use the other value of x

Answer 9

thank you took some time to work it out but got it.