Q) Find the value of ‘k’ for which the quadratic equation (k + 1) x2 – 6 (k + 1) x + 3 (k + 9) = 0, k ≠ – 1 has real and equal roots.
[CBSE 2024 – Series 5 – Set 1]
Ans: Given quadratic equation is: (k + 1) x 2 – 6 (k + 1) x + 3 (k + 9) = 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 = – 6 (k + 1)
and C = 3 (k+ 9)
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:
∴ (- 6 (k + 1))2 – 4 (k + 1) x 3 (k + 9) = 0
∴ 36 (k + 1)2 – 12 (k + 1) (k + 9) = 0
∴ 3 (k + 1)2 – (k + 1) (k + 9) = 0 (divided by 12 on both sides)
∴ (k + 1)[3 (k + 1) – (k + 9)] = 0
∴ (k + 1)(2 k – 6) = 0
∴ k = -1, k = 3
Since it is given that k ≠ – 1, therefore, for k = 3, the given quadratic equation will have real and equal roots
Please press the “Heart” button if you like the solution.
Check: 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 – 6 (k + 1) x + 3 (k + 9) = 0
∴ (3 + 1) x 2 – 6 (3 + 1) x + 3 (3 + 9) = 0
∴ 4 x 2 – 24 x + 36 = 0
∴ x 2 – 6 x + 9 = 0
∴ (x – 3)2 = 0
∴ Roots are: 3 and 3
Since we get two real and equal roots, our answer k = 3 is correct.