Question

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.

Answer 1

(c + 1)x² - 6 (c + 1)x + 3(c + 9) = 0

Comparing the above equation with ax² + bx + c = 0, we get

a = (c + 1), b = - 6(c + 1), c = 3(c + 9)

∴ ∆ = b² – 4ac

= [- 6(c + 1)]² - 4(c + 1) × 3(c + 9)

= 36 (c + 1)² - 12 (c + 1) (c + 9)

= 36 (c² +2c + 1) - 12(c² + 10c + 9)

= 36c² + 72c + 36 - 12c² - 120c - 108

= 24c² − 48c − 72

For real and equal roots, we set ∆ = 0;

24c² − 48c − 72 = 0

Dividing the entire equation by 24 to simplify:

c² − 2c − 3 = 0

Then, factor the quadratic equation

∴ (c - 3)(c + 1) = 0

So, either

∴ c - 3 = 0 ⇒ c = 3

∴ c + 1 = 0 ⇒ c = - 1

However, it is given that c ≠ - 1.

Therefore, the value of c for which the quadratic equation (c + 1) x² - 6(c + 1) x + 3(c + 9) = 0 has real and equal roots is c = 3