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Question

I. 12y2 + 11y – 15 = 0

II. 8x2 – 6x – 5 = 0

A Quantity I > Quantity II Correct Answer Incorrect Answer
B Quantity I ≥ Quantity II Correct Answer Incorrect Answer
C Quantity I < Quantity II Correct Answer Incorrect Answer
D Quantity I ≤ Quantity II Correct Answer Incorrect Answer
E Quantity I = Quantity II or No relation can be established Correct Answer Incorrect Answer

Solution

I. 12y2 + 11y – 15 = 0 12y2 – 9 y + 20y – 15 = 0 3 y(4 y – 3) + 5(4 y – 3) = 0 (3 y + 5) (4 y – 3) = 0 y = 3/4, -5/3 II. 8x2 – 6x – 5 = 0 8x2 – 10x + 4x – 5 = 0 2x(4x – 5) + 1(4x – 5) = 0 (4x  – 5) (2 x+1) = 0 x = 5/4, -1/2 Hence, relationship cannot be established between x and y. Alternate Method: Remember whenever roots of x & y are + & - both, there is no need to find the exact value of roots. In this case, there will be always no relation between x & y. Hence relationship between x & y cannot be established.

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