The smallest value of $|a|$ such that the lines $ x = a+m $, $y = -2 $ and $y = mx$ are concurrent

Question

Question

If the line $ x = a+m $, $y = -2 $ and $y = mx$ are concurrent, the least value of $|a|$ is

(A) $\sqrt{2}$ (B) $2\sqrt{2}$ (C) $2\sqrt{3}$ (D) $3\sqrt{2}$

Solution

Since the lines are concurrent $-2 = m(a +m)$ $\Rightarrow$ $m^2 + am + 2 = 0 .$ Since $m$ is real, $a^2\ge 8,$ $|a| \ge 2\sqrt{2}.$ Hence least value of $|a|$ is $2\sqrt{2}.$

So my question is how $a^2 \ge 8$ was assigned the value. If anyone can please give a hint it would be helpful.

Answer 1

Hint: Do you know what the discriminant of a quadratic function is? In order for a quadratic to have real roots, the discriminant must be non-negative (and it must be positive for it to have distinct real roots).