The Quadratic Formula and the Discriminant: Practice

Match the equation to the template $ax^2 + bx + c = 0$, reading the coefficient of each power.

$a = 1, \qquad b = 5, \qquad c = 6$

The coefficient of $x^2$ is $1$ (an unwritten $1$ still counts), the coefficient of $x$ is $5$, and the constant is $6$.

Read off $a = 1$, $b = -4$, $c = 3$ and substitute with parentheses so the signs behave.

$\Delta = (-4)^2 - 4(1)(3) = 16 - 12 = 4$

A positive discriminant, so the equation has two distinct real roots.

A positive discriminant makes the right side of $\left(x + \tfrac{b}{2a}\right)^2 = \tfrac{\Delta}{4a^2}$ positive, and a positive number has two distinct real square roots.

$\Delta > 0 \iff \text{two distinct real roots}$

Graphically, the parabola crosses the $x$-axis at two points.

The formula is completing the square run once on the general quadratic, carrying the letters $a$, $b$, $c$ the whole way.

$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$

Taking square roots and isolating $x$ from this completed-square line produces the formula. That is why it is a theorem, not a memorized recipe.

With $a = 1$, $b = -5$, $c = 6$, the discriminant is $25 - 24 = 1$.

$x = \frac{5 \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2}$

That gives $x = \tfrac{6}{2} = 3$ and $x = \tfrac{4}{2} = 2$, so the solutions are $2$ and $3$.

With $\Delta = 0$ the completed-square equation becomes $\left(x + \tfrac{b}{2a}\right)^2 = 0$, which holds only at one value of $x$.

$x = -\frac{b}{2a}$

That single value is the double root, and the parabola is tangent to the $x$-axis there.

The formula splits as $-\tfrac{b}{2a} \pm \tfrac{\sqrt{\Delta}}{2a}$, one distance added and one subtracted from a center.

$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$

The center $-\tfrac{b}{2a}$ is the axis of symmetry, so the two roots straddle it as a mirror pair.

A negative discriminant makes the right side of $\left(x + \tfrac{b}{2a}\right)^2 = \tfrac{\Delta}{4a^2}$ negative, but a real square is never negative.

$\Delta < 0 \iff \text{no real roots}$

So no real number solves the equation, and the parabola never meets the $x$-axis.

With $a = 1$, $b = 6$, $c = 8$, the discriminant is $36 - 32 = 4$.

$x = \frac{-6 \pm \sqrt{4}}{2} = \frac{-6 \pm 2}{2}$

That gives $x = \tfrac{-4}{2} = -2$ and $x = \tfrac{-8}{2} = -4$, so the solutions are $-2$ and $-4$.

The two roots differ only by the $\pm\sqrt{b^2 - 4ac}$ piece, so they merge when that piece is zero.

$\sqrt{b^2 - 4ac} = 0 \iff b^2 - 4ac = 0$

When the discriminant is $0$, the formula gives the single value $-\tfrac{b}{2a}$, a double root.

A zero discriminant gives one repeated real root, and the parabola meets the axis only there.

$4x^2 - 4x + 1 = (2x - 1)^2 \ge 0$

The square is never negative, so the graph sits on or above the axis and touches it at the single point $x = \tfrac{1}{2}$, tangent to the axis.

The formula was derived by completing the square on the general quadratic, using only $a \ne 0$.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \qquad a \ne 0$

So it applies to every quadratic, whether or not it factors, and whatever the sign of the discriminant (a negative one simply reports no real solutions).