# 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

• Oct 2nd 2008, 11:42 AM
Sean12345
Hi again civiliam,

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

Hope this helps.
• Oct 2nd 2008, 11:44 AM
icemanfan
This is what I would do:

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

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

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

$z = 2^{1/3}$
• 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 $z^{3/2}$ 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 $z^{1/2}$.

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

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

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