# Solve Equation w/ Rational Exponent.

• Jan 31st 2007, 04:42 PM
Mulya66
Solve Equation w/ Rational Exponent.
I am lost. The Algebra II book I use doesn't explain concepts very well. Does anyone know how to solve this problem as well as clearly explain their chosen steps?

$x^{\frac{1}{2}} -5x^{\frac{1}{4}} +6=0$

• Jan 31st 2007, 05:08 PM
Soroban
Hello, Mulya66!

I hope there's a typo in the problem.
. . Otherwise, I can't solve it.
. . . . . . . . . . . . . . . . . . . . .

Is it supposed to be: . $x^{\frac{2}{3}} -5x^{\frac{1}{3}} + 6 \:=\:0$

Then it factors: . $\left(x^{\frac{1}{3}} - 2\right)\left(x^{\frac{1}{3}} - 3\right) \:=\:0$

And we have: . $\begin{Bmatrix}x^{\frac{1}{3}} - 2 \:=\:0\quad\Rightarrow\quad x^{\frac{1}{3}}\:=\:2\quad\Rightarrow\quad x\,=\,8 \\

• Jan 31st 2007, 05:13 PM
Mulya66
Oops! It's all a typo! It's really $x^{\frac{1}{2}} -5x^{\frac{1}{4}} + 6 \:=\:0$

(Well, that just shows how tired I am...)

Quote:

Originally Posted by Soroban
Hello, Mulya66!

I hope there's a typo in the problem.
. . Otherwise, I can't solve it.
. . . . . . . . . . . . . . . . . . . . .

Is it supposed to be: . $x^{\frac{2}{3}} -5x^{\frac{1}{4}} + 6 \:=\:0$

Then it factors: . $\left(x^{\frac{1}{3}} - 2\right)\left(x^{\frac{1}{3}} - 3\right) \:=\:0$

And we have: . $\begin{Bmatrix}x^{\frac{1}{3}} - 2 \:=\:0\quad\Rightarrow\quad x^{\frac{1}{3}}\:=\:2\quad\Rightarrow\quad x\,=\,8 \\

• Jan 31st 2007, 05:53 PM
Soroban
Hello, Mulya66!

Quote:

$x^{\frac{1}{2}} -5x^{\frac{1}{4}} + 6 \:=\:0$

Use the same technique . . .

It factors: . $\left(x^{\frac{1}{4}} - 2\right)\left(x^{\frac{1}{4}} - 3\right) \:=\:0$

Can you finish it now?

• Jan 31st 2007, 06:11 PM
Mulya66
Quote:

Originally Posted by Soroban

$\left(x^{\frac{1}{4}} - 2\right)\left(x^{\frac{1}{4}} - 3\right) \:=\:0$

$\begin{Bmatrix}x^{\frac{1}{4}} - 2 \:=\:0\quad\Rightarrow\quad x^{\frac{1}{4}}\:=\:2\quad\Rightarrow\quad x\,=\,16 \\