Question

Find the value of '$p$' for which the quadratic equation has equal roots, $x^2-px+9=0$

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 $x^2-px+9=0$ is in the form $a{x}^2+bx+c=0$ where $a=1,b=-p$ and $c=9$.

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

We have, $b^2-4ac=0$

$\Rightarrow (-p{)}^2-(4\times 1\times 9)=0$

$\Rightarrow p^2-36=0$

$\Rightarrow p^2=36$

$\Rightarrow p=\pm \sqrt{36}$

$\Rightarrow p=\pm 6$

Hence, $p=\pm 6$