How would I simplify $\displaystyle (144a^8b^2)^1/2$

I didn't know how to make the 1 over 2 smaller so I tried my bestbut its suppose to be in the top right, thanks if you can help

Printable View

- Jul 13th 2006, 11:08 AMRonishelp simplify please
How would I simplify $\displaystyle (144a^8b^2)^1/2$

I didn't know how to make the 1 over 2 smaller so I tried my bestbut its suppose to be in the top right, thanks if you can help - Jul 13th 2006, 11:15 AMThePerfectHackerQuote:

Originally Posted by**Ronis**

$\displaystyle \sqrt{144a^8b^2}=\sqrt{144}\cdot \sqrt{a^8}\cdot \sqrt{b^2}=12a^4b$

-----

Note: (this is not for you) I know that $\displaystyle \sqrt{b^2}\not = b$, so do not bother me with that. Because in schools terms are positive. - Jul 13th 2006, 11:19 AMRonisQuote:

Originally Posted by**ThePerfectHacker**

- Jul 13th 2006, 11:19 AMQuickQuote:

Originally Posted by**ThePerfectHacker**

BTW. to make an exponent with more than one number all you need to do is wrap it in { }

P.S.*I*would've corrected you hacker :D - Jul 13th 2006, 11:24 AMSoroban
Hello, Ronis!

You have to enclose the entire exponent in braces

. like this: (144a^8b^2)^**{**1/2**}**

You can even write: (144^8b^2)^**{**\frac{1}{2}**}**

Quote:

How would I simplify $\displaystyle (144a^8b^2)^{\frac{1}{2}}$

Each factor is raised to the ½ power: .$\displaystyle (144)^{\frac{1}{2}}(a^8)^\frac{1}{2}}(b^2)^{\frac{ 1}{2}}$

Since $\displaystyle 144 = 12^2$, we have: .$\displaystyle (12^2)^{\frac{1}{2}}(a^8)^{\frac{1}{2}}(b^2)^{ \frac{1}{2}} $

Use that third property of exponents: .$\displaystyle 12^{(2\cdot\frac{1}{2})} \cdot a^{(8\cdot\frac{1}{2})}\cdot b^{(2\cdot\frac{1}{2})} $

. . And we get: .$\displaystyle 12^1\cdot a^4\cdot b^1\;=\;12a^4b$

- Jul 13th 2006, 11:31 AMQuick
Just for reference...

$\displaystyle \sqrt[2]{x}=x^{\frac{1}{2}$

also,

$\displaystyle \sqrt[3]{x}=x^{\frac{1}{3}$

you can remember the rule,

$\displaystyle \sqrt[a]{x}=x^{\frac{1}{a}$

also the rules of powers:

$\displaystyle x^a\times x^b=x^{(a+b)}$

$\displaystyle (x^a)^b=x^{(a\times b)}$

$\displaystyle x^{-a}=\frac{1}{x^a}$