Question

Find the value of $p$ such that the quadratic equation has equal roots $(3p+1)c^2+2(p+1)c+p=0$

Answer 1

We know that while finding the root of a quadratic equation $a{x}^2+bx+c=0$ by quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$,

if $b^2-4ac>0$, then the roots are real and distinct

if $b^2-4ac=0$, then the roots are real and equal and

if $b^2-4ac<0$, then the roots are imaginary.

Here, the given quadratic equation $(3p+1)c^2+2(p+1)c+p=0$ is in the form $a{x}^2+bx+c=0$ where $a=(3p+1),b=2(p+1)=(2p+2)$ and $c=p$.

It is given that the roots are equal, therefore $b^2-4ac=0$

$\Rightarrow (2p+2{)}^2-(4\times (3p+1)\times p)=0\Rightarrow (2p{)}^2+2^2+(2\times 2p\times 2)-4(3p^2+p)=0\Rightarrow (4p^2+4+8p)-12p^2-4p=0\Rightarrow 4p^2+4+8p-12p^2-4p=0\Rightarrow -8p^2+4p+4=0\Rightarrow -4(2p^2-p-1)=0\Rightarrow 2p^2-p-1=0\Rightarrow 2p^2-2p+p-1=0\Rightarrow 2p(p-1)+1(p-1)=0\Rightarrow (2p+1)=0,(p-1)=0\Rightarrow 2p=-1,p=1\Rightarrow p=-\frac{1}{2},p=1$

Hence, $p=-\frac{1}{2}$ or $p=1$