Question

Find the values of p for which the quadratic equation $(2p+1)x^2-(7p+2)x+(7p-3)=0$ has real and equal roots.

Answer 1

$(2p+1)x^2-(7p+2)x+(7p-3)=0$
If they have equal roots, $b^2-4ac=0$

$\Rightarrow (7p+2{)}^2-4(7p-3\left)\right(2p+1)=0$
$\Rightarrow$ (7p+2)² - 4(2p + 1) (7p-3)=0
$\Rightarrow$ 49p² +4+ 28p – 4(14p² +p-3)=0
After simplification, 7p²-24p-16=0
Solving the equation, we get p=4 and -4/7.