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Manager
Joined: 03 Jun 2008
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30 Sep 2008, 10:42
1
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Please find the question in the file attached.
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Gmat_M05.doc [47.5 KiB]

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30 Sep 2008, 10:49
GMBA85 wrote:
Please find the question in the file attached.

= sqrt(x^2+6x+9) - sqrt(y^2-2y+1)
= (x+3) - (y-1)
= x+3-y+1

substitue the value, it results in 87/20
//A//
Attachments

M06.JPG [ 11.73 KiB | Viewed 1076 times ]

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Manager
Joined: 03 Jun 2008
Posts: 136
Schools: ISB, Tuck, Michigan (Ross), Darden, MBS
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30 Sep 2008, 10:51
GMAT TIGER wrote:
GMBA85 wrote:
Please find the question in the file attached.

= sqrt(x^2+6x+9) - sqrt(y^2-2y+1)
= (x+3) - (y-1)
= x+3-y+1

substitue the value, it results in 87/20
//A//

Thats what i gt

but as per gmatclub expl its incorrect.

_________________

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'It's not the ride, it's the rider'

SVP
Joined: 29 Aug 2007
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30 Sep 2008, 11:02
GMBA85 wrote:
GMAT TIGER wrote:
GMBA85 wrote:
Please find the question in the file attached.

= sqrt(x^2+6x+9) - sqrt(y^2-2y+1)
= (x+3) - (y-1)
= x+3-y+1

substitue the value, it results in 87/20
//A//

Thats what i gt

but as per gmatclub expl its incorrect.

since sqrt(x^2+6x+9), and sqrt(y^2-2y+1) both are +ve. so:

sqrt(x^2+6x+9) = (x+3)
sqrt(y^2-2y+1) = (y-1)
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Intern
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30 Sep 2008, 11:04
sqrt(x^2+6x+9) - sqrt(y^2-2y+1)
=sqrt(x + 3)^2 - sqrt(y-1)^2
=sqrt(3/4 + 3)^2 - sqrt(2/5-1)^2
=sqrt(15/4)^2 - sqrt(-3/5)^2
=15/4 - 3/5
=63/20
Manager
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30 Sep 2008, 11:10
aim2010 wrote:
sqrt(x^2+6x+9) - sqrt(y^2-2y+1)
=sqrt(x + 3)^2 - sqrt(y-1)^2
=sqrt(3/4 + 3)^2 - sqrt(2/5-1)^2
=sqrt(15/4)^2 - sqrt(-3/5)^2
=15/4 - 3/5
=63/20

20/10 wats wrong with tiger's approach?
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'It's not the ride, it's the rider'

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30 Sep 2008, 11:16
1
KUDOS
it is a common mistake to assume sqrt(x^2) is x. it could be -x if x <0.

for example in this case sqrt(y-1)^2 would not be y-1, it would be 1-y (since y-1 is -ve)

in simple words, sqrt((-3/5)^2) = 3/5, not -3/5
Director
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30 Sep 2008, 11:30
i got 63/20 by approximation

ended up with sqrt262/4 - 3/5
roughly estimated sqrt262 to be 16
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30 Sep 2008, 13:24
GMBA85 wrote:
aim2010 wrote:
sqrt(x^2+6x+9) - sqrt(y^2-2y+1)
=sqrt(x + 3)^2 - sqrt(y-1)^2
=sqrt(3/4 + 3)^2 - sqrt(2/5-1)^2
=sqrt(15/4)^2 - sqrt(-3/5)^2
=15/4 - 3/5
=63/20

20/10 wats wrong with tiger's approach?

I am also intrested to do so?

since sqrt(x) is always x, not -x.

can anybody explain, why?
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VP
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30 Sep 2008, 16:57
That a good point. Even i took it as y-1. Never thought of considering that it could be negative as well.+1 from me
aim2010 wrote:
it is a common mistake to assume sqrt(x^2) is x. it could be -x if x <0.

for example in this case sqrt(y-1)^2 would not be y-1, it would be 1-y (since y-1 is -ve)

in simple words, sqrt((-3/5)^2) = 3/5, not -3/5
VP
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30 Sep 2008, 17:44
aim2010 wrote:
it is a common mistake to assume sqrt(x^2) is x. it could be -x if x <0.

for example in this case sqrt(y-1)^2 would not be y-1, it would be 1-y (since y-1 is -ve)

in simple words, sqrt((-3/5)^2) = 3/5, not -3/5

Jeez! scary.

y ^ 2 - 2y + 1 can be written as (y-1) ^ 2 or (1-y) ^2.

But why do we have to chose (1-y) over (y-1). I felt like I understood and then I lost the train of thought.

1-y = 3/5 and y-1 = -3/5. In either case the value of square is same.

We should not take the sqrt away before and call it (y-1). The calculation must be done inside the sqrt and then removed.
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30 Sep 2008, 19:25
1
KUDOS
icandy wrote:
aim2010 wrote:
it is a common mistake to assume sqrt(x^2) is x. it could be -x if x <0.

for example in this case sqrt(y-1)^2 would not be y-1, it would be 1-y (since y-1 is -ve)

in simple words, sqrt((-3/5)^2) = 3/5, not -3/5

Jeez! scary.

y ^ 2 - 2y + 1 can be written as (y-1) ^ 2 or (1-y) ^2.

But why do we have to chose (1-y) over (y-1). I felt like I understood and then I lost the train of thought.

1-y = 3/5 and y-1 = -3/5. In either case the value of square is same.

We should not take the sqrt away before and call it (y-1). The calculation must be done inside the sqrt and then removed.

if y - 1 is negative and if you do the calculation and then deduce the square root, you would come up with 1 - y. The point is that there's a small trick here to avoid those calculations: if y - 1 is positive, forget calculations and cancel square with square root. if it is negative, cancel and negate the expression.
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01 Oct 2008, 12:11
This is the place where non-real number is being tested. If I write sqrt(-4) * sqrt(-4), I am multiplying two imaginary numbers.

I had posted my query on the site asking whether anyone has experienced imaginary number related questions in GMAT and I got answer in negation. I think, this question helps me get the answer.
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01 Oct 2008, 15:40
Do u mean it will be sqrt(16) and hence real no?

scthakur wrote:
This is the place where non-real number is being tested. If I write sqrt(-4) * sqrt(-4), I am multiplying two imaginary numbers.

I had posted my query on the site asking whether anyone has experienced imaginary number related questions in GMAT and I got answer in negation. I think, this question helps me get the answer.
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01 Oct 2008, 15:51
GMAT TIGER wrote:
GMBA85 wrote:
Please find the question in the file attached.

= sqrt(x^2+6x+9) - sqrt(y^2-2y+1)
= (x+3) - (y-1)
= x+3-y+1

substitue the value, it results in 87/20
//A//

sqrt(x^2+6x+9) - sqrt(y^2-2y+1) = |x+3| - |y-1|

so |3/4 + 3| - |2/5 - 1| = |15/4| - |-3/5| = 15/4 - 3/5 = 63/20

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