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Solve I. $2 x^{2} - 11 x + 15 = 0$
II. $y^{2} + 4 y + 3 = 0$ . Which relation is correct?

Competition practiceQuadratic Comparison

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

Step-by-Step Solution

I:

(2 x - 5) (x - 3) = 0

→

x = 2.5

, 3. II:

(y + 1) (y + 3) = 0

→

y = - 1

, −3. Positive

x

always exceeds negative

y

, so

x > y

.

Answer: B)

x > y

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