Use factoring to solve the equation for real values of the variables

z^3/2 - 2^1/2 = 0

I think i'm making it harder then it realy is please help

Printable View

- Oct 2nd 2008, 11:06 AMciviliamAnother HW Question Polynomial and Radical Equations
Use factoring to solve the equation for real values of the variables

z^3/2 - 2^1/2 = 0

I think i'm making it harder then it realy is please help - Oct 2nd 2008, 11:42 AMSean12345
Hi again civiliam,

We have $\displaystyle z^{\frac{3}{2}}-2^{\frac{1}{2}}=0$ . Now this leads to $\displaystyle z^{\frac{3}{2}}=2^{\frac{1}{2}}$ . Notice that to get $\displaystyle z$ we need to raise $\displaystyle z^{\frac{3}{2}}$ to the power of $\displaystyle \frac{2}{3}$ since we know that $\displaystyle (a^b)^c=a^{bc}$ . Hence raising both sides of the equation to the power $\displaystyle \frac{2}{3}$ yields $\displaystyle z=2^{\left(\frac{1}{2}\right)\left(\frac{2}{3}\rig ht)}=2^{\frac{1}{3}} $ .

Hope this helps. - Oct 2nd 2008, 11:44 AMicemanfan
This is what I would do:

$\displaystyle z^{3/2} - 2^{1/2} = 0$

$\displaystyle z^{3/2} = 2^{1/2}$

$\displaystyle (z^{3/2})^{2/3} = (2^{1/2})^{2/3}$

$\displaystyle z = 2^{1/3}$ - Oct 2nd 2008, 11:47 AMciviliam
sorry why 2/3?

- Oct 2nd 2008, 11:49 AMicemanfan
- Oct 2nd 2008, 11:50 AMciviliam
ahh I c thank you much!

- Oct 2nd 2008, 11:54 AMciviliam
so i wrote the problem out wrong would I still work it out the same way if the 2^1/2 was z^1/2?

z^3/2 - z^1/2 = 0 - Oct 2nd 2008, 11:57 AMicemanfan
Now you can factor out $\displaystyle z^{1/2}$.

Beginning with $\displaystyle z^{3/2} - z^{1/2} = 0$:

$\displaystyle z^{1/2} \cdot (z - 1) = 0$

z = 0 or 1.

So the problem changes significantly. - Oct 2nd 2008, 12:02 PMciviliam
wow that was simple odd how minor changes to a problem can throw me off so easily thanks again