Answer-I. 2y2 + 11y + 15 = 0 II. 3x2 + 4x - 4= 0

I. 2y2 11y 15 0 2y2 6 y 5 y 15 0 2y y 3 5y 3 0 y - 3, -52 II. 3x2 4x - 4 0 3x2 6 x - 2 x - 4 0 3x x 2 - 2x 2 0 3x - 2 x 2 0 x - 2, 23 Hence, x y. Alternate Method if signs of quadratic equation is ve and ve respectively then the roots of equation will be -ve and -ve. So, roots of first equation y -3, -52 if signs of quadratic equation is ve and -ve respectively then the roots of equation will be -ve and ve. note -ve sign will come in larger root So, roots of second equation x -2, 23 After comparing roots of quadratice eqution we can conclude that x y.

Question

I. 2y2 + 11y + 15 = 0 II.

3x2 + 4x - 4= 0

A If x ≥ y

B If x ≤ y

C If x < y

D If x > y

E If relationship between a and b cannot be established or x = y

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I. 2y2 + 11y + 15 = 0 II.

Solution

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