Discriminant of a quadratic equation

The discriminant of a quadratic $ax^2 + bx + c = 0$ is the quantity $b^2 - 4ac$. It comes from under the square root in the quadratic formula, so its sign decides how many real solutions exist.

There are three cases: $b^2 - 4ac > 0$ gives two real solutions, $b^2 - 4ac = 0$ gives exactly one, and $b^2 - 4ac < 0$ gives no real solutions.

The graph shows why. A parabola with two real solutions crosses the $x$-axis at two points, one with a single solution just touches the axis at one point, and one with no real solutions never reaches the axis.

To use it, first identify $a$, $b$, and $c$, then plug in. For $x^2 - 4x + 3 = 0$, $a = 1$, $b = -4$, $c = 3$, so $b^2 - 4ac = 16 - 12 = 4 > 0$ and there are two solutions.

Many questions run this backwards: they give a condition and ask for a constant. To make $x^2 + mx + 4 = 0$ have exactly one solution, set the discriminant to zero: $m^2 - 4(1)(4) = 0$, so $m^2 = 16$.

How many real solutions does $x^2 + 6x + 5 = 0$ have?
Here $a = 1$, $b = 6$, $c = 5$.
Discriminant $= 6^2 - 4(1)(5) = 36 - 20 = 16$. Since it is positive, there are two solutions.

For what value of $c$ does $x^2 - 4x - c = 0$ have exactly one solution?
Set the discriminant to zero: $(-4)^2 - 4(1)(-c) = 0$.
So $16 + 4c = 0$, giving $c = -4$.

Show that $x^2 + x + 1 = 0$ has no real solutions.
Here $a = 1$, $b = 1$, $c = 1$.
Discriminant $= 1^2 - 4(1)(1) = 1 - 4 = -3$. Since it is negative, there are no real solutions.