# Another HW Question Polynomial and Radical Equations

• Oct 2nd 2008, 11:06 AM
civiliam
Another 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 AM
Sean12345
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 AM
icemanfan
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 AM
civiliam
sorry why 2/3?
• Oct 2nd 2008, 11:49 AM
icemanfan
Quote:

Originally Posted by civiliam
sorry why 2/3?

Because 2/3 is the reciprocal of 3/2. When you raise $\displaystyle z^{3/2}$ to the 2/3 power, the exponent on z becomes 1.
• Oct 2nd 2008, 11:50 AM
civiliam
ahh I c thank you much!
• Oct 2nd 2008, 11:54 AM
civiliam
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 AM
icemanfan
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 PM
civiliam
wow that was simple odd how minor changes to a problem can throw me off so easily thanks again