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04 Nov 2022, 01:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
68% (02:08) correct
32%
(02:29)
wrong
based on 537
sessions
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Not Attempted Yet
If x and y are integers such that \(4x^2 – y^2 +4x + 4y – 3 = 0\), which of the following must be true?
A. y is an even number.
B. y is an odd number.
C. y is positive.
D. y is negative.
E. y is a prime number
A. y is an even number.
B. y is an odd number.
C. y is positive.
D. y is negative.
E. y is a prime number
04 Nov 2022, 04:26
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Bunuel
\(4x^2 – y^2 +4x + 4y – 3 = 0\)
\(4(x^2+x+y) – 3 )= y^2\)
\(Even-Odd=y^2\)
As E-O=Odd, so y is odd.
B
Just for those who may be slightly inquisitive on how do we know other options work or not.
\(4x^2 – y^2 +4x + 4y – 3 = 0\)
\(4x^2 +4x +-y^2+ 4y – 3 +(-1+1)= 0\)
\(4x^2 +4x -1-y^2+ 4y – 4=4x(x+1)-(y-2)^2 = 0\)
\(4x(x+1)+1= (y-2)^2\)……Even+1=(y-2)^2
So y-2 is odd or y is odd. Option A is out and B is in.
4x(x+1)+1 is a square, x could be 3 to get 4*3*4+1 or 49. Thus, \((y-2)^2=49=7^2=(-7)^2\)
Or y could be 9 or -5. All other options C, D and E can be eliminated.
04 Nov 2022, 05:36
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kapil1995
Why should \(4x^2-y^2\) be positive?
Say x is 1 and y is 3
\(4*1^2-3^2=-5\)
Of course their SUM will be positive.
