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I. $x^{2} + 2 x + 1 = 0$
II. $y^{2} + y = 0$ . Decide how $x$ compares with $y$ .

Competition practiceQuadratic Comparison

A $x > y$
B $x$ ≥ $y$
C $x < y$
D $x$ ≤ $y$

Step-by-Step Solution

I:

(x + 1)^{2} = 0

→

x = - 1

. II:

y (y + 1) = 0

→

y = 0, - 1. x = - 1

is never more than

y

, and equals

y

when

y = - 1

, so

x

≤

y

.

Answer: D)

x

≤

y

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