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If x^2 + xy  32 = 0, and x and y are integers, then y could equal eac
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01 Aug 2017, 12:08
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If x^2 + xy  32 = 0, and x and y are integers, then y could equal each of the following except? A) 31 B) 14 C) 2 D) 4 E) 14
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Re: If x^2 + xy  32 = 0, and x and y are integers, then y could equal eac
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01 Aug 2017, 12:17
Y must be a number that is produced from adding the factors of 32.
The factors of 32 are the following: (1,32) (2,16) (4,8) (1,32) (2,16) (4,8). Therefore the only possible combination of these factors that doesn't exist when adding them together is 2.
Answer is C



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If x^2 + xy  32 = 0, and x and y are integers, then y could equal eac
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01 Aug 2017, 12:19
For the quadratic equation ax^2 + bx + c = 0
The sum of the roots is \(\frac{b}{a}\) and the product of the roots is \(\frac{c}{a}\)From our question stem, we know that the product is 32 and sum is y We can have a product of 32 in any of the ways: 1*32,2*16,4*8,4*8,2*16 and 1*32. The sum could be 31,14,4,4,14 and 31. Only Option C(2) is not a valid choice of integer y!
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Re: If x^2 + xy  32 = 0, and x and y are integers, then y could equal eac
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01 Aug 2017, 12:56
I'm sure that it is not the easiest way but anyway: x^2+xy=32 > x(x+y)=32. So we have a multiplication of two integers (as x and y are integers) that yield 32. The factors of 32, that means the numbers that their multiplication by each other can yield 32, are 32, 16, 8, 4, 2, 1. Those numbers and their negative values are the only way to yield 32 by a multiplication of two integers. So in fact there is six options for the multiplication: 32x1; 16x2; 8x4; 32x1; 16x2; 8x4. Now we can plug the values of y and check the answers. If the values of the two number in the multiplication matches to one of the aforementioned cases  y can be the number mentioned in the relevant option.
(A): x(x31)=32. We can easily see that x can be equal to 32 which is a factor of 32. The multiplication is 32x1. (B): x(x14)=32. We can easily see that x can be equal to 16 which is a factor of 32. The multiplication is 16x2. (C): x(x+2)=32. In this case x cannot equal to any of the factors of 32. We can simply see it by looking at the factors 32 or by using the quadratic equation that will give us an irrational number for x (which is invalid). (D): x(x+4)=32. We can see that x can be equal to 4 which is a factor of 32. The multiplication is 8x4. (E): x(x+14)32. We can see that x can be equal to 2 which is a factor of 32 as well. The multiplication is 16x2.
Answer: C



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Re: If x^2 + xy  32 = 0, and x and y are integers, then y could equal eac
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18 Aug 2018, 20:12
Bunuel wrote: If x^2 + xy  32 = 0, and x and y are integers, then y could equal each of the following except?
A) 31 B) 14 C) 2 D) 4 E) 14 We can express the equation as follows: x(x + y) = 32 Since x and y are integers, x and x + y are integer factors of 32. For example, if x = 4, then x + y = 8 which makes y = 4. So choice D is not the answer. If x = 1, then x + y = 32 which makes y = 31. So choice A is not the answer. If x = 2, then x + y = 16 which makes y = 14. So choice B is not the answer. If x = 2, then x + y = 16 which makes y = 14. So choice E is not the answer. Answer: C
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Re: If x^2 + xy  32 = 0, and x and y are integers, then y could equal eac
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