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x2−2xy+y2=0 and xy = 9.
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Updated on: 11 Sep 2018, 14:43
Updated on: 11 Sep 2018, 14:43
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34% (01:22) wrong based on 47 sessions
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\(x^2−2xy+y^2=0\) and \(xy=9\)
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Official answer says D (Cannot be determined) but I strongly believe it is C (Quantities are equal)
Quantity A |
Quantity B |
y |
3 |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Official answer says D (Cannot be determined) but I strongly believe it is C (Quantities are equal)
Originally posted by kruttikaaggarwal on 11 Sep 2018, 12:56.
Last edited by Carcass on 11 Sep 2018, 14:43, edited 1 time in total.
Last edited by Carcass on 11 Sep 2018, 14:43, edited 1 time in total.
Edited by Carcass
Retired Moderator
Joined: 07 Jun 2014
Posts: 4806
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Re: x2−2xy+y2=0 and xy = 9.
[#permalink]
11 Sep 2018, 17:57
11 Sep 2018, 17:57
2
Expert Reply
This is one of the questions you need to pay attention.
Given: \(x^2−2xy+y^2=0\) or \((x-y)^2=0\) or \(x=y\)
Also given that \(xy=9\), substituting value of x in terms of y or \(y^2=9\) Here y can have 2 values 3 and -3.
Hence D is correct
_________________
Given: \(x^2−2xy+y^2=0\) or \((x-y)^2=0\) or \(x=y\)
Also given that \(xy=9\), substituting value of x in terms of y or \(y^2=9\) Here y can have 2 values 3 and -3.
Hence D is correct
_________________
Sandy
If you found this post useful, please let me know by pressing the Kudos Button
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If you found this post useful, please let me know by pressing the Kudos Button
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Re: x2−2xy+y2=0 and xy = 9.
[#permalink]
11 Sep 2018, 18:04
11 Sep 2018, 18:04
sandy wrote:
This is one of the questions you need to pay attention.
Given: \(x^2−2xy+y^2=0\) or \((x-y)^2=0\) or \(x=y\)
Also given that \(xy=9\), substituting value of x in terms of y or \(y^2=9\) Here y can have 2 values 3 and -3.
Hence D is correct
Given: \(x^2−2xy+y^2=0\) or \((x-y)^2=0\) or \(x=y\)
Also given that \(xy=9\), substituting value of x in terms of y or \(y^2=9\) Here y can have 2 values 3 and -3.
Hence D is correct
OHH!! The minus sign, I totally ignored that. Thanks, Sandy! I got confused with the explanation because it stated the combination of y= 9 or 1 as the reasoning, which did not fall in line with the given quadratic equation.
Re: x2−2xy+y2=0 and xy = 9.
[#permalink]
10 Jun 2020, 16:59
10 Jun 2020, 16:59
sandy wrote:
This is one of the questions you need to pay attention.
Given: \(x^2−2xy+y^2=0\) or \((x-y)^2=0\) or \(x=y\)
Also given that \(xy=9\), substituting value of x in terms of y or \(y^2=9\) Here y can have 2 values 3 and -3.
Hence D is correct
Given: \(x^2−2xy+y^2=0\) or \((x-y)^2=0\) or \(x=y\)
Also given that \(xy=9\), substituting value of x in terms of y or \(y^2=9\) Here y can have 2 values 3 and -3.
Hence D is correct
How do you substitute?
_________________
Ever Tried? Ever Failed? No Matter. Try Again. Fail Again. Fail Better!!
Re: x2−2xy+y2=0 and xy = 9.
[#permalink]
10 Jun 2020, 21:25
10 Jun 2020, 21:25
2
Farina wrote:
sandy wrote:
This is one of the questions you need to pay attention.
Given: \(x^2−2xy+y^2=0\) or \((x-y)^2=0\) or \(x=y\)
Also given that \(xy=9\), substituting value of x in terms of y or \(y^2=9\) Here y can have 2 values 3 and -3.
Hence D is correct
Given: \(x^2−2xy+y^2=0\) or \((x-y)^2=0\) or \(x=y\)
Also given that \(xy=9\), substituting value of x in terms of y or \(y^2=9\) Here y can have 2 values 3 and -3.
Hence D is correct
How do you substitute?
x = y
xy = 9
y^2 = 9
Thus, y can be +3 or -3.
Re: x22xy+y2=0 and xy = 9.
[#permalink]
04 Jul 2023, 23:29
04 Jul 2023, 23:29
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