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

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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.

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