Question

Find the value of p so that the quadratic equation px(x-3)+=0 has equal roots.

Answer 1

$p {x}^{2} - 3px = 0 \\ for \: equal \: roots \: {b}^{2} - 4ac = 0 \\ ( { - 3p)}^{2} - 4 \times p \times 1 = 0 \\ 9 {p }^{2} - 4p = 0 \\p(9p - 4) = 0 \\ p = 0 \: \: or \: \: p = \frac{4}{9}$