Find $p$, if quadratic equation $py( y-2)+6=0$ has equal roots.

Given: Quadratic equation $py( y-2)+6=0$ has perfect square.

To do: To find the value of $p$.

Solution:
 

Given: $py( y-2)+6=0$

$\Rightarrow py^2-2py+6=0$

$a=p,\ b=-2p$ and $c=6$

For equal roots $b^2-4ac=0$

$\Rightarrow ( -2p)^2-4\times p\times 6$

$\Rightarrow 4p^2-24p=0$

$\Rightarrow 4p( p-6)=0$

$\Rightarrow 4p=0$ and $p-6=0$

$\Rightarrow p=0$ and $p=6$.

Tutorialspoint

e

Updated on: 10-Oct-2022

77 Views

Kickstart Your Career

Get certified by completing the course

Get Started