Question

Without solving the following quadratic equation, find the value of 'p' for which the given equation has real and equal roots:
$x^2+(p-3)x+p=0$.

Answer 1

For real and equal roots:
$b^2-4ac=0\Rightarrow (p-3{)}^2-4(1\left)\right(p)=0\Rightarrow p^2-6p+9-4p=0\Rightarrow p^2-10p+9=0\Rightarrow p^2-9p-p+9=0\Rightarrow (p-9\left)\right(p-1)=0\Rightarrow p=9orp=1$.