Question

Find the value of p for which the roots are real and equal of quadratic equation (3p+1)x^2+2(p+1)x+p=0

Answer 1

$for \: the \: roots \: to \: be \: real \: and \: equal \: \\ {b}^{2} - 4ac = 0 \\ \\ given \: equation \: - - > (3p + 1) {x}^{2} + 2(p + 1)x + p = 0 \\ \\ a = (3p + 1) \\ b = 2(p + 1) \\ c = p \\ \\ {(2(p + 1))}^{2} - 4(3p + 1)(p) = 0 \\ {(2p + 1)}^{2} - {12p}^{2} - 4p = 0 \\ 4 {p}^{2} + 4p + 1 - 12 {p}^{2} - 4p = 0 \\ - 8 {p}^{2} + 1 = 0 \\ {p}^{2} = \frac{1}{8} \\ p = \sqrt{ \frac{1}{8} }$