

InterviewSolution
1. |
Find the value of p for which the quadratic equation: (p+1)x2 - 6(p+1)x + 3(p + 9) = 0, where, p ≠ -1 has equal roots. Hence, find the roots of the equation. |
Answer» Note: For a quadratic equation, ax2 + bx + c = 0, we have D = b2 – 4ac. If D = 0, then the roots of the quadratic equation are equal. Therefore, (p+1)x2 - 6(p+1)x + 3(p + 9) = 0 will have equal roots when, ⇒ D = 0 ⇒ b2 – 4ac = 0 ⇒ b2 = 4ac Here, b = -6(p+1), a = (p+1) and, c = 3(p+9) ⇒{-6(p + 1)}2 = 4×(p + 1)×3(p + 9) ⇒ 36(p+1)(p+1) = 12(p + 1)(p + 9) ⇒ 3(p+1)=(p + 9) ⇒ 3p + 3 - p - 9 = 0 ⇒ 2p - 6 = 0 ⇒ p = 6/2 ⇒ p = 3 Thus, the value of p is 3 Now, putting the value of p in (p+1)x2 - 6(p+1)x + 3(p + 9) = 0, we get, ⇒ 4x2 - 24x + 36 = 0 On taking 4 common, we get, ⇒ x2 – 6x + 9 = 0 ⇒ (x - 3)2 = 0 ⇒ x = 3 Thus, the root of the given equation is x = 3 |
|