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# (root2*root5)/[(root7-root2-root5)(root7+root2+root5)]

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Math Expert
Joined: 02 Sep 2009
Posts: 52231

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09 Dec 2015, 07:48
10
00:00

Difficulty:

55% (hard)

Question Stats:

68% (01:32) correct 32% (01:30) wrong based on 302 sessions

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$$\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=$$

A. -2
B. -1
C. -1//2
D. 1/2
E. 1

Source: GMATQuantum

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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8789
Location: Pune, India

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27 Jan 2016, 22:15
4
2
Bunuel wrote:
$$\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=$$

A. -2
B. -1
C. -1//2
D. 1/2
E. 1

Source: GMATQuantum

The question tests our use of algebraic identities. There are a few things in the expression that give us hints on how to proceed.

Note the $$\sqrt{2}*\sqrt{5}$$ in the numerator. This reminds us of $$2ab$$ terms in $$(a+b)^2$$ and $$(a-b)^2$$. There are three terms in each factor of the denominator but one term has only positive signs while the other has negative signs too. This reminds us of $$a^2 - b^2$$ expansion.

$$(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5}) =[\sqrt{7}-(\sqrt{2}+\sqrt{5})][\sqrt{7}+(\sqrt{2}+\sqrt{5})]$$

Here $$a = \sqrt{7}$$

$$b = (\sqrt{2}+\sqrt{5})$$

$$(a + b)(a - b) = a^2 - b^2$$

$$[\sqrt{7}-(\sqrt{2}+\sqrt{5})][\sqrt{7}+(\sqrt{2}+\sqrt{5})] = (\sqrt{7})^2 - (\sqrt{2}+\sqrt{5})^2$$

Now use $$(x + y)^2 = x^2 + y^2 + 2xy$$ on $$(\sqrt{2}+\sqrt{5})^2$$

$$(\sqrt{2}+\sqrt{5})^2 = \sqrt{2}^2 + \sqrt{5}^2 + 2*\sqrt{2}*\sqrt{5}= 7 + 2*\sqrt{2}*\sqrt{5}$$

Plugging this value back in the denominator:

$$[\sqrt{7}-(\sqrt{2}+\sqrt{5})][\sqrt{7}+(\sqrt{2}+\sqrt{5})] = 7 - ( 7 + 2*\sqrt{2}*\sqrt{5})$$

$$[\sqrt{7}-(\sqrt{2}+\sqrt{5})][\sqrt{7}+(\sqrt{2}+\sqrt{5})] = - 2*\sqrt{2}*\sqrt{5}$$

Plugging this value of denominator back in the fraction:

$$\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}= \frac{\sqrt{2}*\sqrt{5}}{-2*\sqrt{2}*\sqrt{5}} = \frac{-1}{2}$$

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Karishma
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##### General Discussion
Math Expert
Joined: 02 Aug 2009
Posts: 7198

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09 Dec 2015, 08:23
1
Bunuel wrote:
$$\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=$$

A. -2
B. -1
C. -1//2
D. 1/2
E. 1

Source: GMATQuantum

$$\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=$$...
lets solve the denominator...
$${(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}= (\sqrt{7})^2 - (\sqrt{2}+\sqrt{5})^2$$..
this is equal to $$-2*\sqrt{10}$$...
therefore the fraction becomes $$\sqrt{10}/2*\sqrt{10}$$= -1/2
C
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Manager
Joined: 15 Feb 2015
Posts: 100

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09 Dec 2015, 08:59
chetan2u wrote:
Bunuel wrote:
$$\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=$$

A. -2
B. -1
C. -1//2
D. 1/2
E. 1

Source: GMATQuantum

$$\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=$$...
lets solve the denominator...
$${(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}= (\sqrt{7})^2 - (\sqrt{2}+\sqrt{5})^2$$..
this is equal to $$-2*\sqrt{10}$$...
therefore the fraction becomes $$\sqrt{10}/2*\sqrt{10}$$= -1/2
C

for some crazy reason i am getting -20 as my answer. Is there a different sign in between? Coz, i also got (-2 Sqroot 10) when i simplified the parenthesis.
Math Expert
Joined: 02 Aug 2009
Posts: 7198

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09 Dec 2015, 21:44
rohan89 wrote:
chetan2u wrote:
Bunuel wrote:
$$\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=$$

A. -2
B. -1
C. -1//2
D. 1/2
E. 1

Source: GMATQuantum

$$\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=$$...
lets solve the denominator...
$${(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}= (\sqrt{7})^2 - (\sqrt{2}+\sqrt{5})^2$$..
this is equal to $$-2*\sqrt{10}$$...
therefore the fraction becomes $$\sqrt{10}/2*\sqrt{10}$$= -1/2
C

for some crazy reason i am getting -20 as my answer. Is there a different sign in between? Coz, i also got (-2 Sqroot 10) when i simplified the parenthesis.

if you too have got the denominator as $$-2\sqrt{10}$$, the numerator is$$\sqrt{2}*\sqrt{5}$$=$$\sqrt{10}$$...
strike off the common term $$\sqrt{10}$$..
what is left is -$$\frac{1}{2}$$
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Intern
Joined: 01 Aug 2014
Posts: 43

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16 Jan 2016, 08:33
(√7 - √2 - √5)*( √7+√2+√5) = - (√2+√5 - √7) *( √7+√2+√5) = - [(√2+√5)^2 – 7]
=> -(2 + 5 + 2√10 – 7) = -2√10
√2*√5/ -2√10 = -1/2
Manager
Joined: 03 Aug 2015
Posts: 52
Concentration: Strategy, Technology
Schools: ISB '18, SPJ GMBA '17
GMAT 1: 680 Q48 V35

### Show Tags

27 Jan 2016, 19:43
Bunuel wrote:
$$\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=$$

A. -2
B. -1
C. -1//2
D. 1/2
E. 1

Source: GMATQuantum

Bunuel...Do we have any formula to deal with the denominator?

I solved by multiplying each term....If there is any quick way, please let us know

Thanks,
Arun
Non-Human User
Joined: 09 Sep 2013
Posts: 9416

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24 Dec 2018, 06:32
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Re: (root2*root5)/[(root7-root2-root5)(root7+root2+root5)] &nbs [#permalink] 24 Dec 2018, 06:32
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# (root2*root5)/[(root7-root2-root5)(root7+root2+root5)]

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