Question

Find the value of '$p$' for which the quadratic equatio has equal roots, $(p+1)n^2+2(p+3)n+(p+8)=0$

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 $(p+1)n^2+2(p+3)n+(p+8)=0$ is in the form $a{x}^2+bx+c=0$ where $a=(p+1),b=2(p+3)=(2p+6)$ and $c=(p+8)$.

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

$b^2-4ac=0\Rightarrow (2p+6{)}^2-(4\times (p+1)\times (p+8))=0\Rightarrow (2p{)}^2+6^2+(2\times 2p\times 6)-4\left[p\right(p+8)+1(p+8\left)\right]=0\Rightarrow (4p^2+36+24p)-4(p^2+8p+p+8)=0\Rightarrow (4p^2+36+24p)-4(p^2+9p+8)=0\Rightarrow (4p^2+36+24p)-4p^2-36p-32=0\Rightarrow 4p^2+36+24p-4p^2-36p-32=0\Rightarrow -12p+4=0\Rightarrow -12p=-4\Rightarrow p=\frac{4}{12}\Rightarrow p=\frac{1}{3}$

Hence, $p=\frac{1}{3}$