Question

determine the nature of the roots of the quadratic equation 15x^2+11x+2=0​

Answer 1

Answer:

b^2-4ac =1 >0 so , they are real roots

Answer 2

(i) 15x2 + 11x + 2 = 0 comparing with ax2 + bx + c = 0. Here a = 15, 6 = 11, c = 2. Δ = b2 – 4ac = 112 - 4 x 15 x 2 = 121 – 120 = 1 > 1. ∴ The roots are real and unequal. (ii) x2 – x – 1 = 0, Here a = 1, b = -1, c = -1. Δ = b2 – 4ac = (-1)2 – 4 x 1 x -1 = 1 + 4 = 5 > 0. ∴ The roots are real and unequal. (iii) √2t2 – 3t + 3√2 = 0 Here a = √2, b = -3, c = 3√2 Δ = b2 – 4ac = (-3)2 – 4 x √2 x 3√2 = 9 – 24 = -15 < 0. ∴ The roots are not real. (iv) 9y2 – 6√2y + 2 = 0 a = 9, b = 6√2, c = 2 Δ = b2 – 4ac = (6√2)2 – 4 x 9 x 2 = 36 x 2 – 72 = 72 – 72 = 0 ∴ The roots are real and equal. (v) 9a2 b2 x2 – 24abcdx + 16c2 d2 = 0 Δ = b2 – 4ac = (-24abcd)2 – 4 x 9a2 b2 x 16c2 d2 = 576a2 b2 c2 d2 – 576a2 b2 c2 d2 = 0 ∴