3x^1/3 -16x^1/3 -35=0 I substituted x^1/3 for u and end up getting stuck at 3u^2 -16u-35 What do I do from here?
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Originally Posted by daword177 3x^2/3 -16x^1/3 -35=0 I substituted x^1/3 for u and end up getting stuck at 3u^2 -16u-35 What do I do from here? $\displaystyle 3u^2-16u-35=(3u+5)(u-7)$
That's a quadratic equation (once you add the "= 0") so if you cannot factor, complete the square or use the quadratic formula.
Originally Posted by daword177 3x^1/3 -16x^1/3 -35=0 I substituted x^1/3 for u and end up getting stuck at 3u^2 -16u-35 What do I do from here? Do you mean 3x^2/3 -16x^1/3 -35=0 If so: Your substitution is OK. Use the quadratice formula to solve the equation $\displaystyle 3u^2 -16u-35 = 0$ for u. Afterwards plug in the values of u into the equation $\displaystyle \sqrt[3]{x} = u$ and solve for x.
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