• Jul 13th 2006, 11:08 AM
Ronis
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 AM
ThePerfectHacker
Quote:

Originally Posted by Ronis
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

It means square root,
$\displaystyle \sqrt{144a^8b^2}=\sqrt{144}\cdot \sqrt{a^8}\cdot \sqrt{b^2}=12a^4b$

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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 AM
Ronis
Quote:

Originally Posted by ThePerfectHacker
It means square root,
$\displaystyle \sqrt{144a^8b^2}=\sqrt{144}\cdot \sqrt{a^8}\cdot \sqrt{b^2}=12a^4b$

-----
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.

{s con/con} (lol) thanks =]
• Jul 13th 2006, 11:19 AM
Quick
Quote:

Originally Posted by ThePerfectHacker
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.

Did you think one of us would correct you? Tsk, tsk.

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 AM
Soroban
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 AM
Quick
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:

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

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

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