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# √[2√63 + 2/(8+3√7)] =

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Joined: 17 Jan 2010
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21 Jan 2010, 06:31
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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) $$\sqrt{7}$$
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26 Sep 2010, 12:07
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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

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

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26 Sep 2010, 11:31
1
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√[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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Posts: 106
Location: USA
Schools: IU KSB

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21 Jan 2010, 09:22
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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06 Jul 2010, 11:08
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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06 Jul 2010, 13:19
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!
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Joined: 22 Dec 2009
Posts: 39
Re: √[2√63 + 2/(8+3√7)] =  [#permalink]

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06 Jul 2010, 22:39
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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07 Jul 2010, 06:45
1
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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07 Jul 2010, 12:25
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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26 Sep 2010, 12:15
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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25 Oct 2010, 03:46
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

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11 Feb 2011, 08:41
√[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

√[16] = 4
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Joined: 20 Jun 2011
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15 Jul 2012, 15:06
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 $$\sqrt{16}=4$$

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

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21 Oct 2014, 23:55
$$\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}} = \sqrt{2\sqrt{63}+\frac{2}{8+\sqrt{63}}} = \sqrt{2\sqrt{63}+\frac{2}{8+\sqrt{63}} * \frac{8-\sqrt{63}}{8-\sqrt{63}}} = \sqrt{2\sqrt{63}+16 - 2\sqrt{63}} = \sqrt{16} = 4$$

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18 Jun 2015, 19:18
1
$$(2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}$$

A) $$8+3\sqrt{7}$$
B) $$4+3\sqrt{7}$$
C) 8
D) 4
E) $$\sqrt{7}$$
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18 Jun 2015, 19:35
1
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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18 Jun 2015, 20:31
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)

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

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18 Jun 2015, 22:06
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

Except the answer choice, no other option is even close
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18 Jun 2015, 22:53
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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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18 Jun 2015, 23:03
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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Re: Roots problem &nbs [#permalink] 18 Jun 2015, 23:03

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