petud wrote:
What is the value of (x+y)²?
i) x+y+1 = 0
ii) x² + y² = 1-2xy
An alternate approach is to TEST CASES.
Statement 1: x+y+1 = 0
Case 1: x=1
1 + y + 1 = 0
2 + y = 0
y = -2.
In this case, (x+y)² = (1 + -2)² = 1.
Case 2: x=10
10 + y + 1 = 0
11 + y = 0
y = -11.
In this case, (x+y)² = (10 + -11)² = 1.
Case 3: x=-1/2
-1/2 + y + 1 = 0
1/2 + y = 0
y = -1/2.
In this case, (x+y)² = (-1/2 + -1/2)² = 1.
In every case, (x+y)² = 1.
SUFFICIENT.
Statement 2: x² + y² = 1-2xy
Case 1: x=1
1² + y² = 1 - (2*1*y)
y² + 2y = 0
y(y+2) = 0.
y=0 or y=-2.
If y=0, then, (x+y)² = (1 + 0)² = 1.
If y=-2, then (x+y)² = (1 + -2)² = 1.
Case 2: x=10
10² + y² = 1 - (2*10*y)
y² + 20y + 99 = 0
(y+9)(y+11) = 0.
y=-9 or y=-11.
If y=-9, then, (x+y)² = (10 + -9)² = 1.
If y=-11, then (x+y)² = (10 + -11)² = 1.
In every case, (x+y)² = 1.
SUFFICIENT.
The correct answer is
D.
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