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Quadratic [#permalink]
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30 Sep 2008, 11:42
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Please find the question in the file attached. == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
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Re: Quadratic [#permalink]
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30 Sep 2008, 11:49
GMBA85 wrote: Please find the question in the file attached. = sqrt(x^2+6x+9)  sqrt(y^22y+1) = (x+3)  (y1) = x+3y+1 substitue the value, it results in 87/20 //A//
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Re: Quadratic [#permalink]
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30 Sep 2008, 11:51
GMAT TIGER wrote: GMBA85 wrote: Please find the question in the file attached. = sqrt(x^2+6x+9)  sqrt(y^22y+1) = (x+3)  (y1) = x+3y+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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Re: Quadratic [#permalink]
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30 Sep 2008, 12:02
GMBA85 wrote: GMAT TIGER wrote: GMBA85 wrote: Please find the question in the file attached. = sqrt(x^2+6x+9)  sqrt(y^22y+1) = (x+3)  (y1) = x+3y+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^22y+1) both are +ve. so: sqrt(x^2+6x+9) = (x+3) sqrt(y^22y+1) = (y1)
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Re: Quadratic [#permalink]
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30 Sep 2008, 12:04
sqrt(x^2+6x+9)  sqrt(y^22y+1) =sqrt(x + 3)^2  sqrt(y1)^2 =sqrt(3/4 + 3)^2  sqrt(2/51)^2 =sqrt(15/4)^2  sqrt(3/5)^2 =15/4  3/5 =63/20



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Re: Quadratic [#permalink]
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30 Sep 2008, 12:10
aim2010 wrote: sqrt(x^2+6x+9)  sqrt(y^22y+1) =sqrt(x + 3)^2  sqrt(y1)^2 =sqrt(3/4 + 3)^2  sqrt(2/51)^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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Re: Quadratic [#permalink]
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30 Sep 2008, 12:16
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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(y1)^2 would not be y1, it would be 1y (since y1 is ve)
in simple words, sqrt((3/5)^2) = 3/5, not 3/5



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Re: Quadratic [#permalink]
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30 Sep 2008, 12:30
i got 63/20 by approximation
ended up with sqrt262/4  3/5 roughly estimated sqrt262 to be 16



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Re: Quadratic [#permalink]
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30 Sep 2008, 14:24
GMBA85 wrote: aim2010 wrote: sqrt(x^2+6x+9)  sqrt(y^22y+1) =sqrt(x + 3)^2  sqrt(y1)^2 =sqrt(3/4 + 3)^2  sqrt(2/51)^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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Re: Quadratic [#permalink]
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30 Sep 2008, 17:57
That a good point. Even i took it as y1. 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(y1)^2 would not be y1, it would be 1y (since y1 is ve)
in simple words, sqrt((3/5)^2) = 3/5, not 3/5



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Re: Quadratic [#permalink]
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30 Sep 2008, 18:44
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(y1)^2 would not be y1, it would be 1y (since y1 is ve)
in simple words, sqrt((3/5)^2) = 3/5, not 3/5 Jeez! scary. y ^ 2  2y + 1 can be written as (y1) ^ 2 or (1y) ^2. But why do we have to chose (1y) over (y1). I felt like I understood and then I lost the train of thought. 1y = 3/5 and y1 = 3/5. In either case the value of square is same. We should not take the sqrt away before and call it (y1). The calculation must be done inside the sqrt and then removed.



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Re: Quadratic [#permalink]
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30 Sep 2008, 20:25
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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(y1)^2 would not be y1, it would be 1y (since y1 is ve)
in simple words, sqrt((3/5)^2) = 3/5, not 3/5 Jeez! scary. y ^ 2  2y + 1 can be written as (y1) ^ 2 or (1y) ^2. But why do we have to chose (1y) over (y1). I felt like I understood and then I lost the train of thought. 1y = 3/5 and y1 = 3/5. In either case the value of square is same. We should not take the sqrt away before and call it (y1). 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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Re: Quadratic [#permalink]
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01 Oct 2008, 13:11
This is the place where nonreal 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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Re: Quadratic [#permalink]
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01 Oct 2008, 16:40
Do u mean it will be sqrt(16) and hence real no? scthakur wrote: This is the place where nonreal 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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Re: Quadratic [#permalink]
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01 Oct 2008, 16:51
GMAT TIGER wrote: GMBA85 wrote: Please find the question in the file attached. = sqrt(x^2+6x+9)  sqrt(y^22y+1) = (x+3)  (y1) = x+3y+1 substitue the value, it results in 87/20 //A// sqrt(x^2+6x+9)  sqrt(y^22y+1) = x+3  y1 so 3/4 + 3  2/5  1 = 15/4  3/5 = 15/4  3/5 = 63/20 The answer is B == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.










