Question

${x}^{2} + m(4x + m - 1) + 2$
find the value of m if the roots are real

Answer 1

$\textit{Quadratic}$
$f(x) = x^2 + m(4x+m-1) + 2$
$\textit{Find m, if function f have real roots.}$

$f(x) = x^2 + 4mx+m^2-m + 2$
$D = b^2-4ac$
$\textit{For real roots,}\, D \geq 0$
$0 \leq b^2-4ac$
$0 \leq 16m^2-4(m^2-m+2)$
$0 \leq 4(4m^2-m^2+m-2)$
$0 \leq 4(3m^2+m-2)$
$0 \leq 4(3m-2)(m+1)$
$m \geq \frac{2}{3} \vee m \leq -1$