Algebra, square root simplification question?

Question

I have gotten the problems down to this:

16 sq.root5 and 32 sq.root5

I am not sure if these are the simplified answers ... are they?

If not how do I continue from here?

Do I just multiply the number in front of the sq. root sign by the actual square root?

Thanks for any help in advance,
Jennifer

Answer 1

Jennifer, those two numbers are simplified since √5 is as far as you can go since it's prime. On a calculator, you just multiply the number in front of the square root times the square root of 5 to get some decimal.

If your problem was to multiply 16√5 times 32√5, then you multiply the rational numbers and the irrational numbers:

16√5 * 32√5
= (16*32) (√5*√5)
= 512 * 5
= 2560

Always remember, a square root times itself loses the square root. √x * √x = x

Answer 2

Those are the simplified answers. The way to tell if a number in the square root is simplified is to use prime factorization. If you have more than one of any prime factor, further simplification is required. Five is a prime number, so the answers are fully simplified. However, for example, if you had the square root of 12. The prime factors would be 2*2*3. Therefore, 2 would come out of the square root, and the simplified form would be 2 sqrt 3.

Answer 3

no what you do is multiply 16 and 32 which equals 512(or add, depending on what the problem tells you) and then multiply the square roots of 5 together and its supposed to give you sq.root of 25 then you take out the sq root of 25 which is 5 then multiply five times 512.

Answer 4

You look to see if there are any multiples of square numbers under the square root sign.

In the case of 5 there are none.

if you have sqrt(8), this is sqrt(4*2) = sqrt(4)*sqrt(2)

The point in separating them is be able to square root the square number

sqrt(8) = 2*sqrt(2)

There are no more squares to factor out :)

I hope this helps.

Answer 5

you may upload sq. roots as long as they have the comparable cost interior. then you definately retain the sq. root cost and purely upload the coefficients in front. sqrt(15) + sqrt(15) = 2sqrt(15) occasion: 3 sqrt (6) + 4 sqrt(6) = 7 sqrt (6)

Answer 6

Leave them as they are.
16 sq.root5 and 32 sq.root5

Answer 7

Your done!!