Question

Please help me.
Simplify. http://static.k12.com/bank_packages/files/media/mathml_921c241132f849473af04b5c3b7021882c543809_1.gif
A. http://static.k12.com/bank_packages/files/media/mathml_4ccf2131dc8d54d8c39ccf6849fd0832ebe9b81d_1.gif
B. http://static.k12.com/bank_packages/files/media/mathml_9e1e39e6349cbbc4c29e9511b5ab902696413d8d_1.gif
C. http://static.k12.com/bank_packages/files/media/mathml_86c8bc73597ae952587c0b74fe44014b11f75bda_1.gif
D. http://static.k12.com/bank_packages/files/media/mathml_56b29458fcf2b31555243ca246f4c34b1838a84e_1.gif

Answer 1

What have you tried?

Answer 2

I'm no good at these and I can't figure it out.

Answer 3

Keep dividing the number under the square-root by a prime number over and over again. You do this to identify squares that you can remove from the radical. For example, if I had \[\sqrt{25}\]
I would divide it by 5 and see how many times it went into 25. I would find that it goes into 25 twice, so I could write \[\sqrt{25} = \sqrt{5*5}\]

For a more complex example, let's try \[\sqrt{64}\]

64 divided by 2 is 32.
32 divided by 2 is 16.
16 divided by 2 is 8.
8 divided by 2 is 4.
4 divided by 2 is 2.

Let's count the 2's: we have 6 of them. So we know that \[\sqrt{64} = \sqrt{2*2*2*2*2*2}\]. For each pair of 2's under the radical, we can put ONE 2 outside the radical.
\[\sqrt{64} = \sqrt{2*2*2*2*2*2} = 2\sqrt{2*2*2*2} = 2*2\sqrt{2*2} = 2*2*2 = 8\].

Answer 4

I didn't look at all of your problems, but you can do this for the 1st one. Try this with \[\sqrt{847}\] but see how many times you can divide 847 by 7. If you reach a point where you can no longer divide by seven, try a different prime number. Go until you reach a prime that is not factorable by a whole number.

Answer 5

For a better explanation, read:

http://www.purplemath.com/modules/radicals.htm

Also work the examples as you read them.

Answer 6

When I divide 847 by 7 I get 121, then when I divide that by 7 I get 17.2857142857142. Should I keep dividing?

Answer 7

You should stop at 121 because you can't divide that by 7 and get a whole number result. Now try a different number to divide 121 with; choose one that will result in a whole number answer. Hint: 121 is a perfect square.

Answer 8

Well 121 by 4 is 30.25 and 121 by 5 is 24.2. Also 121 by 2 is 60.5.

Answer 9

Good tries. Go a little higher. This is just something you learn after a while: 121 is 11 times 11.

Answer 10

So you started with 847. You saw that the last number in the "one's place" is a 7 and decided to start dividing by that (this is a good way to start if you can't easily see some other factors. If the number is even, you know you can start with 2....). You found that 847 divided by 7 is 121. Now you found that 121 divided by 11 is 11. So your factors are 7, 11 and 11.

This is the same as:

\[\sqrt{847} = \sqrt{7*11*11}\]

Now, remove what you can out from the radical. Remember, for every two, you can remove one.

Answer 11

...I figured it out. Its \[11\sqrt7 \] Thank you so much for helping me understand this.

Answer 12

Great, no problem! And good job!

Answer 13

Just FYI "for every two, you can remove one" is for square roots. If you go into cuberoots, e.g., \[8^{\frac{1}{3}}\], you would have 2 * 2 * 2 under the radical, and in the case of cuberoots, that would be "for every 3, you can remove one..." and so on.

Answer 14

Okay thanks again @mathbrz