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?
$\displaystyle x^{\frac{1}{2}} -5x^{\frac{1}{4}} +6=0$
Thanks in advance.
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?
$\displaystyle x^{\frac{1}{2}} -5x^{\frac{1}{4}} +6=0$
Thanks in advance.
Hello, Mulya66!
I hope there's a typo in the problem.
. . Otherwise, I can't solve it.
. . . . . . . . . . . . . . . . . . . . .↓
Is it supposed to be: .$\displaystyle x^{\frac{2}{3}} -5x^{\frac{1}{3}} + 6 \:=\:0$
Then it factors: .$\displaystyle \left(x^{\frac{1}{3}} - 2\right)\left(x^{\frac{1}{3}} - 3\right) \:=\:0$
And we have: .$\displaystyle \begin{Bmatrix}x^{\frac{1}{3}} - 2 \:=\:0\quad\Rightarrow\quad x^{\frac{1}{3}}\:=\:2\quad\Rightarrow\quad x\,=\,8 \\
x^{\frac{1}{3}} - 3\:=\:0\quad\Rightarrow\quad x^{\frac{1}{3}}\:=\:3\quad\Rightarrow\quad x\,=\,27 \end{Bmatrix} $
$\displaystyle \begin{Bmatrix}x^{\frac{1}{4}} - 2 \:=\:0\quad\Rightarrow\quad x^{\frac{1}{4}}\:=\:2\quad\Rightarrow\quad x\,=\,16 \\
x^{\frac{1}{4}} - 3\:=\:0\quad\Rightarrow\quad x^{\frac{1}{4}}\:=\:3\quad\Rightarrow\quad x\,=\,81 \end{Bmatrix} $
Thanks! It's very easy now.