Find the values of p for which the quadratic equation ( p + 1 ) x 2 - 6 ( p + 1 ) x + 3 ( p + 9 ) = 0 , p ≠ - 1 has equal roots. Hence find the roots of the equation.

The given equation is ( p + 1 ) x 2 - 6 ( p + 1 ) x + 3 ( p + 9 ) = 0 This is of the form a x 2 + b x + c = 0 where a = p + 1 , b = - 6 ( p + 1 ) and c = 3 ( p + 9 ) ∴ D = b 2 - 4 a c = [ - 6 ( p + 1 ) ] 2 - 4 × ( p + 1 ) × 3 ( p + 9 ) = 36 ( p + 1 ) 2 - 12 ( p + 1 ) ( p + 9 ) = 12 ( p + 1 ) [ 3 ( p + 1 ) - ( p + 9 ) ] = 12 ( p + 1 ) ( 2 p - 6 ) The given equation will have real and equal roots if D = 0 . ∴ 12 ( p + 1 ) ( 2 p - 6 ) = 0 ⇒ p + 1 = 0 o r 2 p - 6 = 0 ⇒ p = - 1 o r p = 3 But, p ≠ - 1 (Given) Thus, the value of p is 3 Putting p = 3 , the given equation becomes 4 x 2 - 24 x + 36 = 0 ⇒ 4 ( x 2 - 6 x + 9 ) = 0 ⇒ ( x - 3 ) 2 = 0 ⇒ x - 3 = 0 ⇒ x = 3 Hence, 3 is the repeated root of this equation.

Find the values of p for which the quadratic equation ( p + 1 ) x 2 - 6 ( p + 1 ) x + 3 ( p + 9 ) = 0 , p ≠ - 1 has equal roots. Hence find the roots of the equation.

The given equation is ( p + 1 ) x 2 - 6 ( p + 1 ) x + 3 ( p + 9 ) = 0 This is of the form a x 2 + b x + c = 0 where a = p + 1 , b = - 6 ( p + 1 ) and c = 3 ( p + 9 ) ∴ D = b 2 - 4 a c = [ - 6 ( p + 1 ) ] 2 - 4 × ( p + 1 ) × 3 ( p + 9 ) = 36 ( p + 1 ) 2 - 12 ( p + 1 ) ( p + 9 ) = 12 ( p + 1 ) [ 3 ( p + 1 ) - ( p + 9 ) ] = 12 ( p + 1 ) ( 2 p - 6 ) The given equation will have real and equal roots if D = 0 . ∴ 12 ( p + 1 ) ( 2 p - 6 ) = 0 ⇒ p + 1 = 0 o r 2 p - 6 = 0 ⇒ p = - 1 o r p = 3 But, p ≠ - 1 (Given) Thus, the value of p is 3 Putting p = 3 , the given equation becomes 4 x 2 - 24 x + 36 = 0 ⇒ 4 ( x 2 - 6 x + 9 ) = 0 ⇒ ( x - 3 ) 2 = 0 ⇒ x - 3 = 0 ⇒ x = 3 Hence, 3 is the repeated root of this equation.