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Question

# I.3x2+8x+4=0 II.4y2−19y+12=0

A

if x > y

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B

if x ≥ y

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C

if x < y

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D

if x ≤ y

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E

if x = y or the relationship cannot be established

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Solution

## The correct option is C if x < y I. 3x2+ 8x + 4 = 0 Or 3x2+ 6x + 2x + 4 = 0 Or 3x (x + 2) + 2(x + 2) = 0 Or (3x + 2) (x + 2) = 0 Therefore x = −23 or -2 II. 4y2- 19y + 12 = 0 Or 4y2- 16y - 3y + 12 = 0 Or 4y (y - 4) - 3(y - 4) = 0 Or (4y - 3) (y - 4) = 0 Therefore y = 34 or 4 So x < y is the answer.

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