Question

Find the value of '$p$' for which the quadratic equation has equal roots, $2y^2-py+1=0$

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

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

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

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

$b^2-4ac=0\Rightarrow (-p{)}^2-(4\times 2\times 1)=0\Rightarrow p^2-8=0\Rightarrow p^2=8\Rightarrow p=\pm \sqrt{8}\Rightarrow p=\pm 2\sqrt{2}$

Hence, $p=\pm 2\sqrt{2}$