Question

b²-4ac=0
$b {}^{2} - 4ac = 0$
​

Answer 1

Step-by-step explanation:

The roots of the equation ax²+bx+c are given by:

x = {-b ± sqrt[b²-4ac]}/2a

If b²-4ac=0, then x=-b/2a, i.e. one root.

If b²-4ac>0, then x=(-b ± k)/2a, i.e. two roots.

If b²-4ac<0, you'd take the square root of a negative number, so no real roots exist.