Question

Find the value(s) of ‘A’ for which the roots of the following equations are real and equal. 

(i) (5k – 6)x² + 2kx + 1 = 0

(ii) kx² + (6k + 2)x + 16 = 0

Answer 1

(i) Here a = 5k – 6 ; b = 2k and c = 1 

Since the equation has real and equal roots 

∆ = 0.

∴ b² – 4ac = 0 

(2k)² – 4(5k – 6) (1) = 0 

4k² – 20k + 24 = 0

(÷ 4) ⇒ k² – 5k + 6 = 0 

(k – 3) (k – 2) = 0

k - 3 = 0 or k – 2 = 0 

k = 3 or k = 2 

The value of k = 3 or 2

(ii) Here a = k, b = 6k + 2; c = 16 

Since the equation has real and equal roots

∆ = 0 

b² – 4ac = 0 

(6k + 2)² – 4(k) (16) = 0 

36k² + 4 + 24k – 4(k) (16) = 0 

36k² – 40k + 4 = 0 

(÷ by 4) ⇒ 9k² – 10k + 1 = 0 

9k² – 9k – k + 1 = 0

9k(k – 1) – 1(k – 1) = 0 

9k (k – 1) -1 (k – 1) = 0 

(k – 1) (9k – 1) = 0 

k – 1 or 9k – 1 = 0 

k = 1 or k = 1/9

The value of k = 1 or 1/9