Math is absolutely my worst subject and I am now struggling to learn math all over again in preparation for my GRE. Can you please help me solve and explain the two following problems?

(√300) ∕ (√12)

(√5) (√2) - √90

Thank you (Rofl)

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- Jan 2nd 2009, 03:32 PMtbeeWorking with square roots - preparing for GRE
Math is absolutely my worst subject and I am now struggling to learn math all over again in preparation for my GRE. Can you please help me solve and explain the two following problems?

(√300) ∕ (√12)

(√5) (√2) - √90

Thank you (Rofl) - Jan 2nd 2009, 03:45 PMLast_Singularity
- Jan 2nd 2009, 04:04 PMtbee
Thanks, how about the second problem. The book answer is -2√10 and I can't figure out how they came up with that since you can only add/subtract when numbers inside square root are identical and have no idea of how to break it down.

- Jan 2nd 2009, 04:42 PMLast_Singularity
$\displaystyle \sqrt{5}\sqrt{2} - \sqrt{90}$

$\displaystyle \sqrt{(2)(5)}- \sqrt{(9)(10)}$

$\displaystyle \sqrt{10} - \sqrt{9} \sqrt{10}$

$\displaystyle \sqrt{10} - 3 \sqrt{10}$

$\displaystyle - 2 \sqrt{10}$ - Jan 10th 2009, 08:12 PMlebanon
- Jan 10th 2009, 10:06 PMGrandadRoots
Hello lebanon

Your first solution above is the answer to the problem $\displaystyle \sqrt{300} - \sqrt{12}$, not $\displaystyle \sqrt{300} / \sqrt{12}$. This is because $\displaystyle \sqrt{300} = \sqrt{100\times 3} = 10\sqrt{3}$ and $\displaystyle \sqrt{12} = \sqrt{4\times 3} = 2\sqrt{3}$. Is that you meant?

Grandad