What can be the values of $m$ in $(m-3)x^2 + (m+2)x + 2m + 1$?

Question

What can be the values of $m$, if $(m-3)x^2 + (m+2)x + 2m + 1$ should be always greater than or equal to zero?

I think we should be using Delta, but I've got no idea how.

Answer 1

Let, \begin{align} y&=ax^2+bx+c\\ &=a\left(x+\dfrac{b}{2a}\right)^2+\left(c-\dfrac{b^2}{4a}\right)\\ &=a\left(x+\dfrac{b}{2a}\right)^2-\left(\dfrac{b^2-4ac}{4a}\right)\\ &=a\left[\left(x+\dfrac{b}{2a}\right)^2-\dfrac{D}{4a^2}\right] \end{align}

From this we can conclude that, when $D\le0$ and $a\ge0$, then only $y\ge0$. Applying this to $y=(m-3)x^2 + (m+2)x + 2m + 1$: \begin{align} D&\le0\\ \implies(m+2)^2-4(m-3)(2m+1)&\leq0\\ \implies7m^2-24m-16&\ge0\\ \implies(7m+4)(m-4)&\ge0\\ \implies m\le-\dfrac47\quad\text{and,}\quad m&\ge4\tag{i} \end{align} Again, $a\ge0\implies m-3\ge0\implies m\ge3.\tag{ii}$

Therefore from (i) and (ii) we get, $m\ge4$.

Answer 2

$m-3>0$ and $(m+2)^2-4(m-3)(2m+1)\leq0$, which gives $m\geq4$.