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**Q)** Find the value of ‘c’ for which the quadratic equation (c + 1) x2 – 6 (c + 1) x + 3 (c + 9) = 0; c ≠ 1 has real and equal roots.

** [CBSE 2024 – Series 4 – Set 1]**

**Ans: **Given quadratic equation is: (c + 1) y^{2} – 6 (c + 1) y + 3 (c + 9) = 0

If we compare to A x^{2} + B x + C = 0, then we can see that

A = (c + 1),

B = – 6 (c + 1)

and C = 3 (c + 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 (c + 1))^{2} – 4( c + 1) x 3(c + 9) = 0

∴ 36 (c + 1)^{2} – 12 (c + 1)(c + 9) = 0

∴ 3 (c^{2}+ 2c + 1) – (c^{2} +10 c + 9) = 0 (divided by 12 on both sides)

∴ 3 c^{2 }+ 6 c + 3 – c^{2} -10c – 9 = 0

∴ 2 c^{2} – 4 c – 6 = 0

∴ c^{2} – 2 c – 3 = 0

∴ c^{2} – 3 c + c – 3 = 0

∴ c (c – 3) + 1 (c – 3) = 0

∴ (c – 3) (c + 1) = 0

∴ c = 3, c = – 1

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

**Therefore, for c = 3, the given quadratic equation will have equal roots**

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**Check:*** **Let’s put the value c = 3 in the given quadratic equation and check if we get 2 equal roots or not:*

*(c + 1) y ^{2} – 6 (c + 1) y + 3 (c + 9) = 0*

*∴ (3 + 1) y ^{2 }*– 6 (3 +1) y + 3 (3 + 9) = 0

*∴ 4 y ^{2 }*– 24 y + 36 = 0

* ∴ y ^{2}*– 6 y + 9 = 0

*∴ (y – 3) ^{2}* = 0

*It gives two equal roots of y = 3. Hence, our answer c = 3 is correct.*