If x + y = -10 and xy = 16, which of the following statements must be

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

If x + y = -10 and xy = 16, which of the following statements must be true?

(I) x – y = -6 --> wrong: \((x-y)^2\) = \((x+y)^2\) -4xy = 10^2-4*16=36 => x-y = +6 or -6

(II) \(x^2 + y^2\) = 68 --> correct: \((x+y)^2\) = x^2+y^2+2xy => 10^2=x^2+y^2+2*16=> x^2+y^2=68

(III) \((x-y)^2\) leaves a remainder of 1 when divided by 7 --> correct: \((x-y)^2\) =36 = 7*5+1


A) I only
B) II only
C) III only
D) I and II
E) II and III --> correct