Math hmwk. :(

Even easier: Reduce the fraction 48/75 to lowers terms first to get 16/25 which is just 4^{2}/5^{2}.

Then take the the square root to get 4/5.

Math hmwk. :(

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Even easier: Reduce the fraction 48/75 to lowers terms first to get 16/25 which is just 4^{2}/5^{2}.

Then take the the square root to get 4/5.

Laihh - Is that √(48/75) or √48/75 ?

If √(48/75), this is equal to √48/√75 . We then look for squares within the numerator and denominator:

√48/√75 = (√16√3)/(√25√3) = (4√3)/(5√3) . Reduce the √3/√3, and you are left with 4/5.

If √48/75, follow the same steps for the numerator to get 4√3/75.

i concur with the above answer

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## Comments

Yes this is a great answer.

Roman, reducing the fraction is smart, and leads us to the answer faster which is why we should always see if we can reduce first but not every problem with evaluating square roots are going to deal with finding the square root of a fraction.

The main principle that needs to be addressed is that we have to look for the perfect squares inside the radical.

To do this we just square a few counting numbers starting from 2 until their squares exceed the biggest number in the radical. Which in this case since we reduced the fraction inside the radical to 16/25, it would be 25.

2

^{2}=4, 3^{2}=9, 4^{2}=16, 5^{2}=25. We can stop here since 25 divides the denominator perfectly, and 16 divides the numerator perfectly. This means that the fraction can now be re-written as follows:v((4

^{2})/(5^{2})), the squareroot undoes the squares and we are left with: 4/5.The reason why the concept kevin expressed is so important is because you won't always be given a fraction.

Here is an example of how this method can be used.

Find v128

Like I stated before, begin by creating a list of perfect squares until you exceed the number inside the radical: 2

^{2}=4, 3^{2}=9, 4^{2}=16, 5^{2}=25, 6^{2}=36, 7^{2}=49, 8^{2}=64, 9^{2}=81, 10^{2}=100, 11^{2}=121, 12^{2}=144, we stop there and exclude 12 since it exceeds 128.Now we may begin to see which is the largest perfect square that divides 128 starting from largest to smallest: Meaning we eliminate 121, 100, and 81 until we arive at 64 which divides 128 into 2.

With that being said we can now re-write 128 as 64·2, and we can rewrite that into 8

^{2}·2.So now everything should look like this: v(8

^{2}·2)The 8 leaves the radical losing its radical, and we are left with 8v2.