Q) Find the value of ‘k’ for which the quadratic equation (k + 1) x ² – 2 (3 k + 1) x + (8 k + 1) = 0 has real and equal roots.

 [CBSE 2024 – Series 4 – Set 2]

Ans: Given quadratic equation is: (k + 1) x ² – 2 (3 k + 1) x + (8 k + 1) = 0

If we compare with the standard quadratic equation A x ² + B x + C = 0, then we can see that

A = (k + 1),

B = – 2 (3 k + 1)

and C = (8 k+ 1)

For a quadratic equation to have real (positive) and equal roots, the discriminant D should be $0$.

or B ² – 4 A C = 0

Substituting the values of A, B and C, we get:

∴ (- 2 (3 k + 1))² – 4 (k + 1) x (8 k+ 1) = 0

∴ 4 (3 k + 1)² – 4 (k + 1) (8 k+ 1) = 0

∴ (3 k + 1)² – (k + 1) (8 k+ 1) = 0      (divided by 4 on both sides)

∴ (9 k ² + 6 k + 1) – (8 k ² + 8 k + k + 1) = 0

∴ (9 k ² + 6 k + 1 – 8 k ² – 9 k – 1) = 0

∴ k ²  – 3 k = 0

∴ k (k -3) = 0

∴ k = 0, k = 3

Since it is given that c ≠ – 1, Hence, c = 3.

Therefore, for k = 0 and k = 3, the given quadratic equation will have real and equal roots

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Check:  Case I (k = 0):  Let’s put the value k = 0 in the given quadratic equation and check if we get 2 real and equal roots or not:

(k + 1) x ² – 2 (3 k + 1) x + (8 k + 1) = 0

∴  (0 + 1) x ² – 2 (3 x 0 + 1) x + (8 x 0 + 1) = 0

∴   x ² – 2 x + 1 = 0

∴   (x – 1) ² = 0

It gives two real and equal roots of x = 1. Hence, our answer k = 0 is correct.

Case II (k = 3): Let’s put the value k = 3 in the given quadratic equation and check if we get 2 real and equal roots or not:

(k + 1) x ² – 2 (3 k + 1) x + (8 k + 1) = 0

∴  (3 + 1) x ² – 2 (3 x 3 + 1) x + (8 x 3 + 1) = 0

∴   4 x ² – 20 x + 25 = 0

∴   (2 x – 5) ² = 0

It gives two real and equal roots of x = . Hence, our answer k = 3 is correct.