Question

Find the value of p in a quadratic equation (3p+1)c^2+2(p+1)c+p=0 has equal roots​

Answer 1

Answer:

Here, the given quadratic equation

(3p+1)c ^2+2(p+1)c+p=0 is in the form

ax^2+bx+c=0 where a=(3p+1),b=2(p+1)=(2p+2) and c=p.

It is given that the roots are equal, therefore

b^2−4ac=0 that is:

b^2−4ac=0

⇒(2p+2)^2 −(4×(3p+1)×p)=0

⇒(2p)^2 +(2×2p×2)+2^2−4(3p^2 +p)=0

⇒(4p^2+4+8p)−12p^2−4p=0

⇒4p^2 +4+8p−12p^2−4p=0

⇒−8p^2+4p+4=0

⇒−4(2p^2 −p−1)=0

⇒2p^2−p−1=0

⇒2p^2−2p+p−1=0

⇒2p(p−1)+1(p−1)=0

⇒(2p+1)=0,(p−1)=0

⇒2p=−1,p=1

⇒p=− 1/2,p=1

Hence, p=− 1/2 or p=1