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If x and y are integers such that 4x^2 y^2 +4x + 4y 3 = 0 04 Nov 2022, 01:30
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B
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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


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B
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If x and y are integers such that 4x^2 y^2 +4x + 4y 3 = 0 04 Nov 2022, 04:26
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Bunuel
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

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\(4x^2 – y^2 +4x + 4y – 3 = 0\)
\(4(x^2+x+y) – 3 )= y^2\)
\(Even-Odd=y^2\)
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.
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If x and y are integers such that 4x^2 y^2 +4x + 4y 3 = 0 04 Nov 2022, 05:36
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kapil1995
\(4x^2 – y^2 +4x + 4y – 3 = 0\)

\(4x^2 – y^2\)....\(x^2\) and \(y^2\) will be always positive

so , \(4x^2 – y^2\) will be positive.

Now to become whole equation = 0 , This part must be negative 4x + 4y – 3

So, y must be negative,

Option D
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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.
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If x and y are integers such that 4x^2 y^2 +4x + 4y 3 = 0 10 Nov 2022, 07:09
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4x^2 + 4x + 4y - y^2 - 3 = 0 (Note E + E = E)
E E E

Even - 3 - y^2 = 0 (Note: Even - Odd (which is 3) = Odd)

Odd - y^2 = 0

Hence, y^2 has to be odd so that LHS = RHS; i.e. 0.

Hope this is clear!
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If x and y are integers such that 4x^2 y^2 +4x + 4y 3 = 0 03 Jan 2023, 04:42
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Took a slightly different approach to the ones above.

Simply equation down to:
(2x+y)(2x-y)+4(x+y)=3

Note that 3 is odd, and 4(x+y) is always even. We have an odd sum, so it must be odd+even. This means (2x+y)(2x-y) needs to be odd.
As 2x is always even, the only way for 2x+y and 2x-y to be odd is for y to be odd. There are no other possible combinations, so y must be odd.

Therefore, B.
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If x and y are integers such that 4x^2 y^2 +4x + 4y 3 = 0 06 Jan 2023, 06:51
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B. y is an odd number.

1. Simplify the equation
\(4x^2 –y^2 + 4x +4y – 3=0 \)
\(4x^2 - y^2 + 4 (x+y) - 3 = 0\)

2. Both \(4x^2\) and \(4(x+y)\) will turn out to be even number b/c anything multiplied by even number, is even:

\(Some even number - y^2 + some even number - 3 = 0\)

Or simply, the sum of two even numbers is also even

\(Some even number - 3 - y^2 = 0\)

3. If we reduce an even number by odd number (3 in this case), it'll always return an odd:

\(Odd number - y^2 = 0\)

Now, y^2 must be an odd number to set the equation to 0; hence, odd*odd -> returns an odd.
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If x and y are integers such that 4x^2 y^2 +4x + 4y 3 = 0 27 Dec 2024, 01:54
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Bunuel
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


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\(4x^2 – y^2 +4x + 4y – 3 = 0\)
\(4x^2 + 4x + 1 - y^2 + 4y - 4 = 0\)
\({(2x + 1)}^{2} - {(y - 2)}^{2} = 0\)
\(((2x+1) - (y-2))*((2x+1) + (y-2)) = 0\)

\(a^2 - b^2 = (a - b)*(a + b)\)

\((2x - y + 3)*(2x + y - 1) = 0\)

2x - y + 3 = 02x + y - 1 = 0
2x - y = -32x + y = 1
Even - Odd = OddEven + Odd = Odd

Hence, y has to be a odd number.

Answer: B
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If x and y are integers such that 4x^2 y^2 +4x + 4y 3 = 0 03 Mar 2026, 14:27
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