OG 12th edition sqrt problem

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OG 12th edition sqrt problem [#permalink]

12 Nov 2009, 06:04

#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!

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sqrt_problem.png [ 182.14 KiB | Viewed 1651 times ]

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Re: OG 12th edition sqrt problem [#permalink]

12 Nov 2009, 06:17

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

Hope it helps.
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Re: OG 12th edition sqrt problem [#permalink]

12 Nov 2009, 08:24

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?

Thanks again.

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Re: OG 12th edition sqrt problem [#permalink]

12 Nov 2009, 08:42

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capable2 wrote:

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:

16*20+8*32=16*20+8*2*16=16*(20+8*2)=16*(20+16)=16*36.

Back to the root. We'll have \sqrt{16*36}=4*6=24

Generally \sqrt{a*b}=\sqrt{a}*\sqrt{b}

Hope it's clear.
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Re: OG 12th edition sqrt problem [#permalink]

29 Jan 2012, 21:15

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.

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Re: OG 12th edition sqrt problem [#permalink]

30 Jan 2012, 02:23

joshuaRome wrote:

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.

For more on this topic check Number theory chapter of Math Book (part about roots): math-number-theory-88376.html

Tough problems on exponents and roots (with detailed solutions) to practice:
PS: tough-and-tricky-exponents-and-roots-questions-125956.html
DS: tough-and-tricky-exponents-and-roots-questions-125967.html

Hope it helps.
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Re: OG 12th edition sqrt problem [#permalink]

30 Jan 2012, 02:52

joshuaRome wrote:

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.

Can we write\sqrt{4} as \sqrt{2+2}? Sure.
But \sqrt{2+2} \neq \sqrt{2} + \sqrt{2}

If addition/subtraction are under the square root, it is fine. But you cannot separate the added terms and cannot give them separate square roots.

Therefore, \sqrt{16+20} = \sqrt{36}

But, \sqrt{16+20} \neq \sqrt{16} + \sqrt{20}

For basic theory of roots, check out: http://www.veritasprep.com/blog/2011/08 ... -the-gmat/
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Re: OG 12th edition sqrt problem [#permalink] 30 Jan 2012, 02:52

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OG 12th edition sqrt problem

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