# Math Help - extracting roots

1. ## extracting roots

I don't know how to simplify these tasks:

1. √2 + √32 =

2. 3*
√44 + 7*√99 =

2. Originally Posted by makaveli89
I don't know how to simplify these tasks:

1. √2 + √32 =

2. 3*
√44 + 7*√99 =
I'll do the first one, and then can you finish the second one?

$\sqrt{2} + \sqrt{32}$

You can turn that into:

$\sqrt{2} + \sqrt{16\cdot 2}$

which can be broken up into:

$\sqrt{2} + \sqrt{16}\sqrt{2}$

Now the square root of 16 is 4 so we now have:

$\sqrt{2} + 4\sqrt{2}$

So then we just add them

$5\sqrt{2}$

so: $\sqrt{2} + \sqrt{32} = 5\sqrt{2}$

You do almost the same thing for the other one, so you should try it out yourself.

3. 3*√44 + 7*√99 =

3*
√11 * √4 + 7* √11 * √9

= 3*2
√11 + 7*3 √11

= 27 *
√11

4. Originally Posted by makaveli89
3*√44 + 7*√99 =

3*
√11 * √4 + 7* √11 * √9

= 3*2
√11 + 7*3 √11

= 27 *
√11
Yep that's exactly right.

5. ## Complex but easy

1. √2 + √32
√2 + (√2 * √2 * √2 * √2 * √2)
√2 + 4√2
5√2

2. 3 = √3 * √3
7 = √7 * √7
Therefore-
√396 + √4851 = 27 * √11

6. Originally Posted by carl_patel
2. 3 = √3 * √3
7 = √7 * √7
Therefore-
√396 + √4851 = 27 * √11
The gap in this is too large. A solution if it is complete should comprise a number of steps each obviously following from its predecessors, or with an explanation in words of why it does.

Also we try to discourage people from giving complete answers to questions, or answers to all the parts of a question, hints, and/or partial answers are preferred.

RonL