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Intern  Joined: 17 Jan 2010
Posts: 23
\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}}  [#permalink]

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3
4 00:00

Difficulty:   25% (medium)

Question Stats: 78% (01:55) correct 22% (02:26) wrong based on 367 sessions

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$$\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}}$$ =

A) $$8+3\sqrt{7}$$
B) $$4+3\sqrt{7}$$
C) 8
D) 4
E) $$\sqrt{7}$$
##### Most Helpful Expert Reply
Math Expert V
Joined: 02 Sep 2009
Posts: 55150
Re: square root problem  [#permalink]

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2
5
udaymathapati wrote:
√[2√63 + 2/(8+3√7)] =
A. 8 + 3√7
B. 4 + 3√7
C. 8
D. 4
E. √7

$$\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}}=\sqrt{6\sqrt{7}+\frac{2}{8+3\sqrt{7}}*\frac{8-3\sqrt{7}}{8-3\sqrt{7}}}=\sqrt{6\sqrt{7}+\frac{2(8-3\sqrt{7})}{64-63}}=\sqrt{6\sqrt{7}+16-6\sqrt{7}}=\sqrt{16}=4$$.

Answer: D.

Please read and apply when posting such questions: writing-mathematical-symbols-in-posts-72468.html
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√[2√63 + 2/(8+3√7)] =  [#permalink]

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1
5
√[2√63 + 2/(8+3√7)] =

A. 8 + 3√7
B. 4 + 3√7
C. 8
D. 4
E. √7
##### General Discussion
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Re: \sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}}  [#permalink]

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1
4
rahulms wrote:
$$\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}$$ =

A) $$8+3\sqrt{7}$$
B) $$4+3\sqrt{7}$$
C) 8
D) 4
E) $$\sqrt{7}$$

$$\sqrt{2\sqrt{63}+\frac{2}{(8+3\sqrt{7})}\frac{(8-3\sqrt{7})}{(8-3\sqrt{7})}$$

$$\sqrt{{2\{3\sqrt{7}})+\frac{2 (8-3\sqrt{7})}{(64-63)}}$$

$$\sqrt{{2\{3\sqrt{7}})+{2 (8)-(2)(3\sqrt{7})}$$

$$\sqrt{16}$$

$$4$$
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GMAT 1: 680 Q48 V34 √[2√63 + 2/(8+3√7)] =  [#permalink]

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1
2
√[2√63 + 2/(8+3√7)] =

A. 8 + 3√7
B. 4 + 3√7
C. 8
D. 4
E. √7
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Re: √[2√63 + 2/(8+3√7)] =  [#permalink]

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2
This is a fairly simple problem but one that requires combination of a few concepts. Here's my simplification:

You first identify that $$\sqrt{63}= 2*3\sqrt{7}$$ and then multiply the numerator and denominator of the second term with $$8-3\sqrt{7}$$ to rationalize it.

Rationalizing the second term yields the following result:

$$\frac{16-6\sqrt{7}}{64-63}$$ = $$\frac{16-6\sqrt{7}}{1}$$ = $$16-6\sqrt{7}$$

The first term is $$2*3\sqrt{7}$$ = $$6\sqrt{7}$$

When you put them together, the expression inside the overall square root symbol becomes:

$$6\sqrt{7}$$ - $$6\sqrt{7} + 16 = 16$$

The $$6\sqrt{7}$$ cancels out leaving you with $$\sqrt{16} = 4$$

Thus, answer is D. Hope this helps!
Intern  Joined: 22 Dec 2009
Posts: 38
Re: √[2√63 + 2/(8+3√7)] =  [#permalink]

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I dont understand,,

why do you multiply with $$8-3\sqrt{7}$$ but not $$8+3\sqrt{7}$$..

And how did you get $$\frac{16-6\sqrt{7}}{64-63}$$

thanks...
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Re: √[2√63 + 2/(8+3√7)] =  [#permalink]

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gmatJP wrote:
I dont understand,,

why do you multiply with $$8-3\sqrt{7}$$ but not $$8+3\sqrt{7}$$..

And how did you get $$\frac{16-6\sqrt{7}}{64-63}$$

thanks...

It's called rationalization gmatJP.

When you multiply $$8+3\sqrt{7}$$ with $$8+3\sqrt{7}$$ you get $$(8+3\sqrt{7})^2$$ which can only be simplified as $$64+ 2*8*3\sqrt{7}+63$$ on the denominator which will not help us much in the process of simplification. The rationale behind multiplying it with $$8-3\sqrt{7}$$ is because it'll be of the form $$(a+b)(a-b) = a^2-b^2$$ which gives us $$(8^2)-(3\sqrt{7})^2$$ = $$64 - (3)^2*(\sqrt{7})^2$$ = $$64-63 = 1$$ which literally makes the denominator inconsequential to the simplification process and helps us cancel terms.

So, now when we multiply the numerator of the second term (2) with $$8-3\sqrt{7}$$ we get $$2*(8-3\sqrt{7})$$ = $$16-6\sqrt{7}$$

So we get $$\frac{16-6\sqrt{7}}{64-63}$$ as the final product. Hope this is clear.
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Re: √[2√63 + 2/(8+3√7)] =  [#permalink]

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Hussain15 wrote:
√[2√63 + 2/(8+3√7)] =

A. 8 + 3√7
B. 4 + 3√7
C. 8
D. 4
E. √7

There is another way to do this without rationalizing it, which I'm not so sure the GMAT even expects you to do.

√[2√63 + 2/(8+3√7)

Multiply the terms: 2√63 + 2/(8+3√7) -> (16√63+6√(63*7)+2)/(8+3√7). Now simplify.

(16*3√7+6*3*7+2)/(8+3√7) -> (16*3√7+128)/(8+3√7)

Now, notice if we take out the 16 from the top we get -> 16*(3√7+8)/(8+3√7) --> and the "messy portion gets cancelled out and we are left with 16. Don't forget that we still have a √ left

√16 = 4 and thus D is our answer.
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Posts: 176
Re: square root problem  [#permalink]

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In case you have forgotten how to work these, I found a good site on conjugates:

http://www.regentsprep.org/Regents/math ... lesson.htm Look at situation 2 and 3.
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Re: Square root  [#permalink]

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In these type of question u multiply the nominator and denominator with conjugate...Conjugate of 8 +3sqrt(7) is 8 -3sqrt(7)

sqrt[2sqrt(63) + 2 { 8 - 3sqrt(7)}/{64 - 63}]
=sqrt[2sqrt(63) +16 - 2sqrt(63)] =4

Answer is 4.

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Re: square root problem  [#permalink]

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√[2√63 + 2/(8+3√7)]

48√7 + 2 * 3 * 3 * √7 * √7 + 2
√[-----------------------------------------]
8 + 3√7

48√7 + 128
√[---------------]
8 + 3√7

16(3√7 + 8)
√[----------------]
8 + 3√7

√ = 4
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Posts: 44
Re: square root problem  [#permalink]

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udaymathapati wrote:
√[2√63 + 2/(8+3√7)] =
A. 8 + 3√7
B. 4 + 3√7
C. 8
D. 4
E. √7

Here is my approach that I think work good for problems like these and on the actual test:

$$\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}}$$.

Since $$\sqrt{64}=8$$
then take: $$\sqrt{2*8}$$

And since this part will be very small
$$\frac{2}{8+3\sqrt{7}}$$.

I just take: $$\sqrt{2*8+ something$$

And get $$\sqrt{16}=4$$

Answer: D.
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Re: √[2√63 + 2/(8+3√7)] =  [#permalink]

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$$\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}} = \sqrt{2\sqrt{63}+\frac{2}{8+\sqrt{63}}} = \sqrt{2\sqrt{63}+\frac{2}{8+\sqrt{63}} * \frac{8-\sqrt{63}}{8-\sqrt{63}}} = \sqrt{2\sqrt{63}+16 - 2\sqrt{63}} = \sqrt{16} = 4$$

Answer = D
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Roots problem  [#permalink]

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$$(2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}$$

A) $$8+3\sqrt{7}$$
B) $$4+3\sqrt{7}$$
C) 8
D) 4
E) $$\sqrt{7}$$
Math Expert V
Joined: 02 Aug 2009
Posts: 7671
Re: Roots problem  [#permalink]

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anurag356 wrote:
$$(2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}$$

A) $$8+3\sqrt{7}$$
B) $$4+3\sqrt{7}$$
C) 8
D) 4
E) $$\sqrt{7}$$

$$(2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}$$
mutliply $$\frac{2}{8+3\sqrt{7}}$$with $$[fraction]8-3\sqrt{7}$$ in noth num and denominator
we get$$2(8-3\sqrt{7})=16-6\sqrt{7}$$
$$2\sqrt{63}=6\sqrt{7}$$
therefore ans is $$\sqrt{16}=4$$
ans D
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Re: Roots problem  [#permalink]

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anurag356 wrote:
$$(2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}$$

A) $$8+3\sqrt{7}$$
B) $$4+3\sqrt{7}$$
C) 8
D) 4
E) $$\sqrt{7}$$

You can solve it easily if you use approximation.

You have two terms:
$$2\sqrt{63}$$
$$\frac{2}{8+3\sqrt{7}}$$

The term $$\frac{2}{8+3\sqrt{7}}$$ is 2 divided by a relatively much larger number so its value will be small (in decimals 0.1 types). That is much smaller than the other term $$2\sqrt{63}$$ because $$\sqrt{63}$$ is almost 8. So the combined effect of these terms is:

2* almost 8 + very small term = 16 = 4^2 (approximately)

$$(4^2)^{1/2}= 4$$

Answer (D)
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Re: Roots problem  [#permalink]

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anurag356 wrote:
$$(2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}$$

A) $$8+3\sqrt{7}$$
B) $$4+3\sqrt{7}$$
C) 8
D) 4
E) $$\sqrt{7}$$

Here is a different and Much faster approach of Pure Aptitude (and less of maths)

We can solve every such question with the approximation

Now we know that $$\sqrt{63}$$ is very close to$$\sqrt{64}$$=8

i.e. we can take $$\sqrt{63}$$= 7.9 or almost 8 for approximation
i.e. $$2*\sqrt{63} = 2*8$$ = approx. $$16$$

Now we know that $$\sqrt{7}$$ is very close to $$\sqrt{9}$$=3
i.e. we can take $$\sqrt{7}= 2.8$$ or almost $$3$$ for approximation

Now, $$8+3\sqrt{7} = 8+3*3 = 17$$ approximately

Now, $$\frac{2}{8+3\sqrt{7}} = 2/17$$ = approx. $$1/8$$ = $$0.125$$ approximately

i.e. $$(2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}$$ = $$(16+0.125)^{1/2}$$ = Approx. 4

Answer: Option

Except the answer choice, no other option is even close
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Re: Roots problem  [#permalink]

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When you see a square root in a denominator of a fraction, you need to get it out of the denominator - that's a required simplification in math. When that root is in a sum or subtraction, you need to use the difference of squares pattern, as below. Note also we can factor a perfect square from √63, so we can also simplify that root:

\begin{align} \left( 2 \sqrt{63} + \frac{2}{8 + 3\sqrt{7}} \right)^{\frac{1}{2}} &= \left( 2 \sqrt{9} \sqrt{7} + \left( \frac{2}{8 + 3\sqrt{7}} \right) \left( \frac{8 - 3\sqrt{7}}{8 - 3\sqrt{7}} \right) \right)^{\frac{1}{2}} \\ &= \left( 6 \sqrt{7} + \frac{16 - 6\sqrt{7}}{8^2 - (9)(7)} \right)^{\frac{1}{2}} \\ &= \left(6 \sqrt{7} + \frac{16 - 6\sqrt{7}}{1} \right)^{\frac{1}{2}} \\ &= (16)^{\frac{1}{2}} \\ &= 4 \end{align}

This is identical to the approach chetan2u used above, but I post it in case the additional detail helps anyone. Estimation also works here, but I wouldn't advise relying on that strategy, since it will only work when the answer choices are sufficiently far apart.
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Re: Roots problem  [#permalink]

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in GMAT, answer choices are sufficiently far apart in the questions of roots (atleast my little 10 years of experience say so about such questions). But yes, We all are entitled to have different opinions. (y)
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