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



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