Question

Determine k such that the quadratic equation x² + 7(3 + 2k) – 2x (1 + 3k) = 0 has equal roots :
(a) 2, 7
(b) 7, 5
(c). 2
(d) -10/9​

Answer 1

Answer:

The equation is x2 – x(2) (1 + 3k) + 7 (3 + 2&) = 0

The roots are real and equal ⇒ ∆ = 0 (i.e.,) b2 – 4ac = 0

Here a = 1, b = -2 (1 + 3k), c = 7(3 + 2k)

So b2 – 4ac = 0 ⇒ [-2 (1 + 3k)]2 – 4(1) (7) (3 + 2k) = 0

(i.e.,) 4 (1 + 3k)2 – 28 (3 + 2k) = 0

(÷ by 4) (1 + 3k)2 – 7(3 + 2k) = 0

1 + 9k2 + 6k – 21 – 14k = 0

9k2 – 8k – 20 = 0

(k – 2)(9k + 10) = 0

k - 2 = 0 or 9k + 10 = 0

k = 2 or k = -10/9