Rational Exponents

**illeatyourxface** · July 16th 2009, 08:31 PM

i dont understand what to do when is it like this

(9+16)^1/2

my answer was 7? is that correct???

and what if it is like this

9^1/2 + 16^1/2

i also got 7???

i have more questions...

**Stroodle** · July 16th 2009, 08:40 PM

Hi.

$(9+16)^{\frac{1}{2}}=25^{\frac{1}{2}}=\sqrt{25}=5$

$9^{\frac{1}{2}}+16^{\frac{1}{2}}=\sqrt{9}+\sqrt{16 }=3+4=7$

**AlephZero** · July 16th 2009, 08:42 PM

9+16=25. So $25^{1/2}=\sqrt{25}=5.$

Your answer to the second question was correct.

EDIT: Too slow...

**illeatyourxface** · July 16th 2009, 08:48 PM

hi, thank you :]

my other questions are...

27^2/3

and i got (27^1/3)^2 = 3^2 = 9?

andd the next one is

x^5/8 divided by X ^1/6

i dont know how to start that one...

and my other question is...

x^2/3(x^1/4 -x)

i dont knowww :[

**Stroodle** · July 16th 2009, 08:57 PM

When dividing you subtract the indices and when multiplying you add the indices (first the indices need to have a common denominator):

$x^{\frac{5}{8}}\div x^{\frac{1}{6}}= x^{\frac{15}{24}}\div x^{\frac{4}{24}}$ $=x^{\frac{11}{24}}$

For the second one you need to add the indices:

$x^{\frac{2}{3}}(x^{\frac{1}{4}}-x)= x^{\frac{8}{12}}\times x^{\frac{3}{12}}-x^{\frac{2}{3}}\times x^{\frac{3}{3}}=x^{\frac{11}{12}}-x^{\frac{5}{3}}$

Does that make sense?

**AlephZero** · July 16th 2009, 09:03 PM

Your first answer is correct.

For the second part, find common denominators among the fractional exponents:

$\frac{x^{5/8}}{x^{1/6}}=\frac{x^{15/24}}{x^{4/24}}=x^{11/24}.$

Similarly for the third question, which I'll let you try yourself.

**illeatyourxface** · July 16th 2009, 09:19 PM

For the second one you need to add the indices:

$x^{\frac{2}{3}}(x^{\frac{1}{4}}-x)= x^{\frac{8}{12}}\times x^{\frac{3}{12}}-x^{\frac{2}{3}}\times x^{\frac{3}{3}}=x^{\frac{11}{12}}-x^{\frac{5}{3}}$

Does that make sense?[/quote]

im a little confused how do you get x^11/12?

**Stroodle** · July 16th 2009, 09:28 PM

$x^{\frac{2}{3}}(x^{\frac{1}{4}}-x)$ is the same as $x^{\frac{8}{12}}(x^{\frac{3}{12}}-x)$

So as you're multiplying $x^{\frac{8}{12}}$ by $x^{\frac{3}{12}}$ the indices added together $=\frac{11}{12}$

Similarly, as $x^{\frac{2}{3}}(x^{\frac{1}{4}}-x)$ is the same as $x^{\frac{2}{3}}(x^{\frac{1}{4}}-x^{\frac{3}{3}})$ you get $x^{\frac{5}{3}}$ as the second term.

**illeatyourxface** · July 16th 2009, 09:31 PM

ohhhh okayyyy thanks so much

i have another simliar question

(x^1/3 + x^-1/3)^2

how would you do that?

**Stroodle** · July 16th 2009, 09:40 PM

$(x^{\frac{1}{3}}+x^{-\frac{1}{3}})^2$

$=\left( x^{\frac{1}{3}}+\frac{1}{x^{\frac{1}{3}}}\right)\l eft( x^{\frac{1}{3}}+\frac{1}{x^{\frac{1}{3}}}\right)$

$=(x^{\frac{1}{3}})^2+2(x^{\frac{1}{3}})(x^{-\frac{1}{3}})+(x^{-\frac{1}{3}})^2$

$=x^{\frac{2}{3}}+2+\frac{1}{x^{\frac{2}{3}}}$

**illeatyourxface** · July 16th 2009, 09:52 PM

so the x's cant be simplified anymore after that???

and i have another question, wanted to see if my answer was correct...

cube root 125

and the answer is 5?

**Stroodle** · July 16th 2009, 10:01 PM

Nope. That's fully simplified.

Your answer is correct. $\sqrt[3]{125}=5$ as $5\times 5\times 5=125$

You should check out the exponent section in this site Practical Algebra Lessons , it explains them really well.

**yeongil** · July 16th 2009, 10:52 PM

Originally Posted by Stroodle

$(x^{\frac{1}{3}}+x^{-\frac{1}{3}})^2$

$=x^{\frac{2}{3}}+x^{-\frac{2}{3}}$

$=x^{\frac{2}{3}}+\frac{1}{x^{\frac{2}{3}}}$

Whoa, that's not right. You have to FOIL it:
$(x^{\frac{1}{3}}+x^{-\frac{1}{3}})^2$
$\begin{aligned}<br /> &= (x^{\frac{1}{3}})^2 + 2(x^{\frac{1}{3}})(x^{-\frac{1}{3}}) + (x^{-\frac{1}{3}})^2 \\<br /> &= x^{\frac{2}{3}} + 2 + x^{-\frac{2}{3}}<br /> \end{aligned}$

01

**Stroodle** · July 16th 2009, 11:48 PM

Whoa, yeah.

Sorry illeatyourxface. Silly mistake.

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