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Given I. $x^{2} - 4 x - 21 = 0$
II. $y^{2} - 19 y + 88 = 0$ , how does $x$ compare with $y$ ?

Competition practiceQuadratic Comparison

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

Step-by-Step Solution

I:

(x - 7) (x + 3) = 0

, so

x = 7

or −3. II:

(y - 8) (y - 11) = 0

, so

y = 8

or 11. The largest

x

(7) is less than the smallest

y

(8), so

x < y

.

Answer: B)

x < y

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