Question

Find the integer value of p for which the quadratic equation (p + 1)$x^2-4\sqrt{3}(p-1)x+3(p+1)=0$, p $\ne$ -1 has equal roots.

Answer 1

$(p+1)x^2-4\sqrt{3}(p-1)x+3(p+1)=0$

$\Rightarrow$ Here, $a=(p+1),b=-4\sqrt{3}(p-1),c=3(p+1)$

$\Rightarrow$ It is given that equation has equal roots.

$\therefore$ $b^2-4ac=0$

$\Rightarrow$ $[-4\sqrt{3}(p-1){]}^2-4(p+1\left)3\right(p+1)=0$

$\Rightarrow$ $48(p^2-2p+1)-12(p^2+2p+1)=0$

$\Rightarrow$ $48p^2-96p+48-12p^2-24p-12=0$

$\Rightarrow$ $36p^2-120p+36=0$

$\Rightarrow$ $3p^2-10p+3=0$

$\Rightarrow$ $3p^2-9p-p+3=0$

$\Rightarrow$ $3p(p-3)-1(p-3)=0$

$\Rightarrow$ $(p-3)(3p-1)=0$

$\Rightarrow$ $p-3=0$ and $3p-1=0$

$\therefore$ $p=3$ and $p=\frac{1}{3}$