# simple question ... just can't figure it out

• Apr 11th 2009, 08:42 AM
lidstrompl
simple question ... just can't figure it out
Solove for y:

y^(2/3) = 2 - y^4

SHOW THE STEPS PLZ
• Apr 11th 2009, 08:51 AM
TKHunny
Why do you believe it can be solved?

What methods have you for your attempt?

What have you tried?

Just looking at it, I think I'd try y = 1 and y = -1.
• Apr 11th 2009, 10:24 AM
stapel
Quote:

Originally Posted by lidstrompl
Solove for y:

y^(2/3) = 2 - y^4

Let x = y^(1/3), so y^(2/3) = x^2 and y^4 = y^(12/3) = x^12.

Substituting, you should get x = 2 - x^12, so you are trying to find the solutions of x^12 + x^2 - 2 = 0, which are the x-intercepts of f = x^12 + x^2 - 2.

Plug x^12 + x^2 - 2 into your graphing calculator to "see" where the two real-values zeroes are. Divide out the two rational zeros (which duplicate the two rational zeroes of the original equation).

You should be left with an equation in x^10 which has no real zeroes.
• Apr 11th 2009, 01:56 PM
lidstrompl
Quote:

Originally Posted by TKHunny
Why do you believe it can be solved?

What methods have you for your attempt?

What have you tried?

Just looking at it, I think I'd try x = 1 and x = -1.

ye, just noticed that as well. Just didn't want to get messy with unnecessary algebra. 1 and -1 in the solution is actually all i need for my problem. Thanks.