# Math Help - help simplify please

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

2. 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.

3. Originally Posted by ThePerfectHacker
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.
{s con/con} (lol) thanks =]

4. 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

5. 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: . $12^1\cdot a^4\cdot b^1\;=\;12a^4b$

6. 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:

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

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

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