Solove for y:

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

SHOW THE STEPS PLZ

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- Apr 11th 2009, 08:42 AMlidstromplsimple 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 AMTKHunny
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 AMstapel
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 PMlidstrompl