I. 2y2 + 11y + 15 = 0 II.

Solution

I. 2y² + 11y + 15 = 0 2y² + 6 y + 5 y + 15 = 0 2y (y + 3) + 5(y + 3) = 0 y = - 3, -5/2 II. 3x² + 4x - 4= 0 3x² + 6 x - 2 x - 4 = 0 3x (x + 2) - 2(x + 2) = 0 (3x - 2) (x + 2) = 0 x = - 2, 2/3 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, -5/2 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, 2/3 After comparing roots of quadratice eqution we can conclude that x > y.