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OG 12th edition sqrt problem [#permalink]
12 Nov 2009, 05:04

#32 on page 156 has a rather simple sqrt problem, which I am able to solve using a brute force method for finding the sqrt. However, I'd like to understand the book's solution for finding the solution more quickly, but don't quite follow it. Any help would be appreciated.

I've attached an image showing the book's solution. I'm wondering how I am able to drop the 16 as one of the root multiples of 256 (the product of 8*32).

Appreciate any help!

Attachments

sqrt_problem.png [ 182.14 KiB | Viewed 3157 times ]

Re: OG 12th edition sqrt problem [#permalink]
12 Nov 2009, 05:17

Expert's post

The solution in book is quite strange. Generally \(\sqrt{a+b}\), DOES NOT equal to \(\sqrt{a}+\sqrt{b}\). Example: \(\sqrt{16+16}=\sqrt{32}=4*\sqrt{2}\) and NOT \(\sqrt{16}+\sqrt{16}=4+4=8\).

Back to the question:

As I understand the problem is:

\(\sqrt{16*20+8*32}\) if yes then: \(\sqrt{16*(20+8*2)}=\sqrt{16*36}=4*6=24\)

Re: OG 12th edition sqrt problem [#permalink]
12 Nov 2009, 07:24

Argh. Sorry, there is a typo in my image - you are correct that it should be \(\sqrt{16}*\sqrt{36}\) and not \(\sqrt{16}+\sqrt{36}\).

I am embarrassed that I still do not understand how (8*32) becomes (8*2)? 8*32 is 256, the root of which is 16. So by my thinking, that would leave me with \(\sqrt{(16)(20)+(16)(16)}\). How do you get \(\sqrt{(16)(20)+(8)(2)}\) from that?

Re: OG 12th edition sqrt problem [#permalink]
12 Nov 2009, 07:42

1

This post received KUDOS

Expert's post

capable2 wrote:

Argh. Sorry, there is a typo in my image - you are correct that it should be \(\sqrt{16}*\sqrt{36}\) and not \(\sqrt{16}+\sqrt{36}\).

I am embarrassed that I still do not understand how (8*32) becomes (8*2)? 8*32 is 256, the root of which is 16. So by my thinking, that would leave me with \(\sqrt{(16)(20)+(16)(16)}\). How do you get \(\sqrt{(16)(20)+(8)(2)}\) from that?

Thanks again.

The red part is not correct we'll get not the expression you wrote but: \(\sqrt{16*(20+8*2)}\). Look at the brackets.

And here is how:

Let's just forget about the square root for a moment. We have:

\(16*20+8*32\) no need to multiply \(8\) by \(32\), it's a long way, though still correct. Just take \(16\) from the brackets:

Re: OG 12th edition sqrt problem [#permalink]
30 Jan 2012, 01:23

Expert's post

joshuaRome wrote:

Hello everyone,

I am having some difficulty with this problem similar to the author of this post.

My problem (and possibly flawed logic):

It is my understanding that \sqrt{2}+\sqrt{2}=2\sqrt{2} and that adding square roots produces the wrong answer.

to say that \sqrt{2}+\sqrt{2}=\sqrt{4}=2 is incorrect from what i see. (1.4appx+1.4appx is not 2)

problem 32 OG:

sqrt{ 16 (20 + 8 * 2)}

sqrt{ 16 (20 + 16)} okay we factored out 16, makes sense to me...

sqrt{16 (36)}

i don't see how we can add two square roots together. square root of 20 + square root 16 is 8.47 not 6.

Thank you for any responses! I know I must have made a mistake but sometimes a different perspective can help.

The point is that we are not adding square root of 20 and square root of 16, we are adding 20 and 16 under the SAME square root. Below it he solution to the problem from OG:

\(\sqrt{16*20+8*32}=?\) --> factor out 16: \(\sqrt{16(20+8*2)}=\sqrt{16*36}=4*6=24\).

Re: OG 12th edition sqrt problem [#permalink]
15 Nov 2014, 15:45

Bunuel wrote:

joshuaRome wrote:

Hello everyone,

I am having some difficulty with this problem similar to the author of this post.

My problem (and possibly flawed logic):

It is my understanding that \sqrt{2}+\sqrt{2}=2\sqrt{2} and that adding square roots produces the wrong answer.

to say that \sqrt{2}+\sqrt{2}=\sqrt{4}=2 is incorrect from what i see. (1.4appx+1.4appx is not 2)

problem 32 OG:

sqrt{ 16 (20 + 8 * 2)}

sqrt{ 16 (20 + 16)} okay we factored out 16, makes sense to me...

sqrt{16 (36)}

i don't see how we can add two square roots together. square root of 20 + square root 16 is 8.47 not 6.

Thank you for any responses! I know I must have made a mistake but sometimes a different perspective can help.

The point is that we are not adding square root of 20 and square root of 16, we are adding 20 and 16 under the SAME square root. Below it he solution to the problem from OG:

\(\sqrt{16*20+8*32}=?\) --> factor out 16: \(\sqrt{16(20+8*2)}=\sqrt{16*36}=4*6=24\).

I just have one question, wouldn't there still be a 1 left over when you factor out the 16? I'm just thinking if you were to reverse the operation, what would the 16 get multiplied into? _________________

"Popular opinion is the greatest lie in the world"-Thomas Carlyle

Re: OG 12th edition sqrt problem [#permalink]
17 Nov 2014, 11:23

Expert's post

Bigred2008 wrote:

Bunuel wrote:

joshuaRome wrote:

Hello everyone,

I am having some difficulty with this problem similar to the author of this post.

My problem (and possibly flawed logic):

It is my understanding that \sqrt{2}+\sqrt{2}=2\sqrt{2} and that adding square roots produces the wrong answer.

to say that \sqrt{2}+\sqrt{2}=\sqrt{4}=2 is incorrect from what i see. (1.4appx+1.4appx is not 2)

problem 32 OG:

sqrt{ 16 (20 + 8 * 2)}

sqrt{ 16 (20 + 16)} okay we factored out 16, makes sense to me...

sqrt{16 (36)}

i don't see how we can add two square roots together. square root of 20 + square root 16 is 8.47 not 6.

Thank you for any responses! I know I must have made a mistake but sometimes a different perspective can help.

The point is that we are not adding square root of 20 and square root of 16, we are adding 20 and 16 under the SAME square root. Below it he solution to the problem from OG:

\(\sqrt{16*20+8*32}=?\) --> factor out 16: \(\sqrt{16(20+8*2)}=\sqrt{16*36}=4*6=24\).

I just have one question, wouldn't there still be a 1 left over when you factor out the 16? I'm just thinking if you were to reverse the operation, what would the 16 get multiplied into?

You mean it should be \(\sqrt{16(20*1+8*2)}=\sqrt{16*36}=4*6=24\)? Yes, but 20*1 = 20, so we can omit 1 there.

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