help simplify please

**Ronis** · July 13th 2006, 11:08 AM

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

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

**ThePerfectHacker** · July 13th 2006, 11:15 AM

Originally Posted by Ronis

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

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

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Note: (this is not for you) I know that $\sqrt{b^2}\not = b$ , so do not bother me with that. Because in schools terms are positive.

**Ronis** · July 13th 2006, 11:19 AM

Originally Posted by ThePerfectHacker

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

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

{s con/con} (lol) thanks =]

**Quick** · July 13th 2006, 11:19 AM

Originally Posted by ThePerfectHacker

Note: (this is not for you) I know that $\sqrt{b^2}\not = b$ , 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

**Soroban** · July 13th 2006, 11:24 AM

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}}

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

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

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

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

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

**Quick** · July 13th 2006, 11:31 AM

Just for reference...

$\sqrt[2]{x}=x^{\frac{1}{2}$
also,
$\sqrt[3]{x}=x^{\frac{1}{3}$
you can remember the rule,
$\sqrt[a]{x}=x^{\frac{1}{a}$

also the rules of powers:

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

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

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

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