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I. $x^{2} + 3 x + 2 = 0$
II. $y^{2} - 5 y + 4 = 0$ . How are $x$ and $y$ related?

Competition practiceQuadratic Comparison

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

Step-by-Step Solution

I:

(x + 1) (x + 2) = 0

→

x = - 1

, −2. II:

(y - 1) (y - 4) = 0

→

y = 1

, 4. Every

x

is negative and every

y

positive, so

x < y

.

Answer: A)

x < y

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