# Thread: Working with square roots - preparing for GRE

1. ## Working 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

2. Originally Posted by tbee
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
Note that $\sqrt{x} \sqrt{y} = \sqrt{xy}$, so for example, $\sqrt{5} \sqrt{2}=\sqrt{10}$

And $\sqrt{x}/\sqrt{y} = \sqrt{x/y}$, so for example, $\sqrt{300}/ \sqrt{12} = \sqrt{300/12}$

3. 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.

4. $\sqrt{5}\sqrt{2} - \sqrt{90}$

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

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

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

$- 2 \sqrt{10}$

5. Originally Posted by tbee
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
the solution is:
=10 redical 3 / 2 redical 3
=5

redical 10 - 3 redical 10
= -2 redical 10

6. ## Roots

Hello lebanon
Originally Posted by lebanon
the solution is:
=10 redical 3 / 2 redical 3
=8 redical 3

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