algebra equation

**Jshizza1228** · January 28th 2009, 06:12 PM

solve the equation by introducing a substitution that transforms the equation to quadratic form........ x^2/3 + 4x^1/3 = 0

**mr fantastic** · January 28th 2009, 06:24 PM

Originally Posted by Jshizza1228

solve the equation by introducing a substitution that transforms the equation to quadratic form........ x^2/3 + 4x^1/3 = 0

Substitute $u = x^{1/3}$ .

**Soroban** · January 28th 2009, 06:25 PM

Hello, Jshizza1228!

Solve the equation by introducing a substitution that transforms
the equation to quadratic form: . $x^{\frac{2}{3}}+ 4x^{\frac{1}{3}} \:= \:0$

We have: . $\left(x^{\frac{1}{3}}\right)^2 + 4\left(x^{\frac{1}{3}}\right) \:=\:0$

Let $z \:=\:x^{\frac{1}{3}}$

. . and we have: . $z^2 + 4z \:=\:0$

Got it?

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