rheam25 wrote:
If x + y = -10 and xy = 16, which of the following statements must be true?
(I) x – y = -6
(II) \(x^2 + y^2\) = 68
(III) \((x-y)^2\) leaves a remainder of 1 when divided by 7
A) I only
B) II only
C) III only
D) I and II
E) II and III
Solution:
Squaring the first equation, we have:
x^2 + y^2 + 2xy = 100
Since xy = 16, we have:
x^2 + y^2 + 32 = 100
x^2 + y^2 = 68
We see that II is true.
Notice that (x - y)^2 = x^2 + y^2 - 2xy = 68 - 32 = 36. So III is also true.
(Note: We don’t need to check statement I since there is no choice that says “All” or “I, II and III”.)
Alternate Solution:Since xy = 16, neither x nor y can equal 0; so let’s write y = 16/x and substitute in x + y = -10:
x + 16/x = -10
x^2 + 16 = -10x
x^2 + 10x + 16 = 0
(x + 2)(x + 8) = 0
x = -2 or x = -8
If x = -2, then y = -8 and if x = -8, then y = -2. Let’s analyse the given statement in the light of this information:
Statement I: x - y = -6
If x = -2 and y = -8, then x - y = -2 - (-8) = 6; so x - y = -6 is not necessarily true.
Statement II: x^2 + y^2 = 68
If x = -2 and y = -8, then x^2 + y^2 = (-2)^2 + (-8)^2 = 4 + 64 = 68.
If x = -8 and y = -2, then x^2 + y^2 = (-8)^2 + (-2)^2 = 64 + 4 = 68.
Statement II is always true.
Statement III: (x - y)^2 leaves a remainder of 1 when divided by 7
We notice that x - y can equal 6 or -6; so (x - y)^2 equals 36 in either case. Since 36 leaves a remainder of 1 when divided by 7, statement III is always true.
Answer: E _________________
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