The quadratic formula and the discriminant

Use the quadratic formula, the axis of symmetry formula and the discriminant to find zeroes, vertices and the number of x-intercepts of any parabola.

The quadratic formula

Completing the square on the general quadratic $ax^2 + bx + c = 0$ gives a formula that solves every quadratic equation. It is on the HSC reference sheet, but you should know it by heart.

Quadratic formula

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

Axis of symmetry

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

Example 1●

Solve $2x^2 - 3x - 4 = 0$. Give exact answers and answers to 2 decimal places.

$a = 2$, $b = -3$, $c = -4$

Write down a, b and c first, including their signs. This avoids most errors.

$b^2 - 4ac = 9 - 4(2)(-4) = 9 + 32 = 41$

Work out the part under the square root on its own.

$x = \dfrac{3 \pm \sqrt{41}}{4}$

$x \approx 2.35$ or $x \approx -0.85$

The discriminant

The expression under the square root, $\Delta = b^2 - 4ac$, is called the because it discriminates between the three possible cases.

Discriminant

$$\Delta = b^2 - 4ac$$

Discriminant	Roots of $ax^2+bx+c=0$	Graph
$\Delta > 0$	Two distinct real roots	Crosses the $x$-axis twice
$\Delta = 0$	One real root (a double root)	Touches the $x$-axis at the vertex
$\Delta < 0$	No real roots	Does not meet the $x$-axis

If $\Delta$ is a perfect square ($0, 1, 4, 9, \ldots$) and $a$, $b$, $c$ are rational, the roots are rational, so the quadratic factors nicely.

Discriminant explorer

Try this

Find values of $a$, $b$, $c$ that make $\Delta = 0$. Where is the vertex?
With $a = 1$ and $b = 2$, find the largest $c$ that still gives zeroes.
Can you make $\Delta < 0$ with $a > 0$ and $c < 0$? Why not?

Always positive, always negative

If $\Delta < 0$ the parabola never meets the $x$-axis, so it is entirely above it or entirely below it:

$a > 0$ and $\Delta < 0$: always positive (positive definite).
$a < 0$ and $\Delta < 0$: always negative (negative definite).

Example 2●●

For what values of $k$ does $x^2 + 4x + k = 0$ have two distinct real roots?

Need $\Delta > 0$.

Two distinct real roots need a positive discriminant.

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

$16 - 4k > 0$, so $4k < 16$ and $k < 4$.

Example 3●●●

Show that $y = -2x^2 + 3x - 5$ is negative for all $x$.

$\Delta = 3^2 - 4(-2)(-5) = 9 - 40 = -31$

$\Delta < 0$, so there are no zeroes.

No zeroes, so the graph is entirely on one side of the x-axis.

$a = -2 < 0$, so the parabola is concave down. Hence it is negative definite: $y < 0$ for all $x$.

Concave down plus no zeroes means the whole graph is below the axis.

Flashcards

Quadratic formula

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

Discriminant

$\Delta = b^2 - 4ac$

Δ > 0

Two distinct real roots

Δ = 0

One double root; vertex on the x-axis

Δ negative

No real roots

Positive definite

$a > 0$ and $\Delta < 0$: always above the x-axis.

Practice

Find the discriminant of $x^2 - 5x + 2$.
$3x^2 + 2x + 1 = 0$ has:
Find $k$ so that $x^2 - 6x + k = 0$ has exactly one root.
The axis of symmetry of $y = 3x^2 - 12x + 1$ is:
$y = x^2 + x + 1$ is:
Solve $x^2 - 2x - 1 = 0$. Give the larger root to 2 decimal places.
The discriminant of $4x^2 - 4x + 1$ is:
Find the positive value of $k$ for which $x^2 + kx + 16 = 0$ has equal roots.