Zeros of Quadratics and the Discriminant

The zeros of a quadratic function are the $x$-values where its graph crosses the $x$-axis. They tell you when a ball hits the ground, when a profit is zero, or where an arch meets the road. A quadratic can have two zeros, one, or none, and a quick calculation called the discriminant tells you which before you solve.

Key ideas

Zeros, roots, and x-intercepts

For a function $f(x)$, the zeros are the solutions of $f(x) = 0$. The same numbers are called the roots of the equation $f(x) = 0$, and they’re the $x$-intercepts of the graph.

Three forms of a quadratic

Form	Equation	What it shows
standard	$f(x) = ax^2 + bx + c$	$y$-intercept $c$
factored	$f(x) = a(x - r)(x - s)$	zeros $r$ and $s$
vertex	$f(x) = a(x - h)^2 + k$	vertex $(h, k)$

In every form, $a \gt 0$ means the parabola opens up and $a \lt 0$ means it opens down.

Finding zeros

• Factoring: write $f(x)$ in factored form, then set each factor equal to $0$.
• Quadratic formula: for $ax^2 + bx + c = 0$,

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

The formula always works, even when the quadratic doesn’t factor nicely.

The discriminant

The expression under the square root, $D = b^2 - 4ac$, is the discriminant:

• $D \gt 0$: two zeros (the square root gives two different answers).
• $D = 0$: one zero (the parabola’s vertex just touches the $x$-axis).
• $D \lt 0$: no real zeros (you can’t take the square root of a negative number).

The discriminant tells you which picture you’ll get.

Reasoning from vertex form

You can often count zeros without any calculation. If the vertex is below the $x$-axis and the parabola opens up, it must cross twice. If the vertex is above the axis and the parabola opens up, it never reaches the axis.

Worked examples

Example 1: By factoring

Find the zeros of $f(x) = x^2 - 2x - 15$.

Solution. Find two numbers that multiply to $-15$ and add to $-2$: they’re $-5$ and $3$.

$$x^2 - 2x - 15 = (x - 5)(x + 3) = 0$$

So $x = 5$ or $x = -3$.

Example 2: With the quadratic formula

Find the zeros of $f(x) = 2x^2 + 3x - 4$. Give exact answers and decimals to two places.

Solution. Here $a = 2$, $b = 3$, $c = -4$. The discriminant is $D = 3^2 - 4(2)(-4) = 9 + 32 = 41$.

$$x = \frac{-3 \pm \sqrt{41}}{4}$$

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

Example 3: Counting zeros

How many zeros does each function have?

(a) $f(x) = 3x^2 - 5x + 4$ $\qquad$ (b) $g(x) = 4x^2 - 12x + 9$ $\qquad$ (c) $h(x) = x^2 + 6x + 2$

Solution.

(a) $D = (-5)^2 - 4(3)(4) = 25 - 48 = -23$. Negative, so no zeros.

(b) $D = (-12)^2 - 4(4)(9) = 144 - 144 = 0$. One zero. (It’s $x = \tfrac{3}{2}$, since $4x^2 - 12x + 9 = (2x - 3)^2$.)

(c) $D = 6^2 - 4(1)(2) = 36 - 8 = 28$. Positive, so two zeros.

Example 4: Finding an unknown coefficient

For which values of $k$ does $f(x) = x^2 + kx + 9$ have exactly one zero?

Solution. One zero means $D = 0$:

$$k^2 - 4(1)(9) = 0 \quad\Rightarrow\quad k^2 = 36 \quad\Rightarrow\quad k = 6 \text{ or } k = -6$$

Check: $x^2 + 6x + 9 = (x + 3)^2$ and $x^2 - 6x + 9 = (x - 3)^2$, each with one zero. ✓

Common mistakes

Using the formula before the equation equals $0$. For $x^2 + 3x = 10$, first rewrite it as $x^2 + 3x - 10 = 0$, so $c = -10$.

Squaring a negative $b$ wrongly. If $b = -5$, then $b^2 = (-5)^2 = 25$, not $-25$.

Dividing only part of the numerator by $2a$. The whole expression $-b \pm \sqrt{b^2 - 4ac}$ is divided by $2a$.

Forgetting the $\pm$. Without it, you’ll find only one of the two zeros.

Mixing up the signs in factored form. $f(x) = (x - 5)(x + 3)$ has zeros $5$ and $-3$: each zero is the number that makes its bracket $0$.

Practice

1. (Warm-up) Find the zeros of $f(x) = (x - 4)(2x + 1)$.

Solution

$x - 4 = 0$ gives $x = 4$, and $2x + 1 = 0$ gives $x = -\tfrac{1}{2}$.

2. (Warm-up) Find the discriminant of $f(x) = 2x^2 - 7x + 3$, and say how many zeros $f$ has.

Solution

$D = (-7)^2 - 4(2)(3) = 49 - 24 = 25$. Positive, so two zeros.

3. (Warm-up) Find the zeros of $f(x) = x^2 + x - 12$ by factoring.

Solution

$x^2 + x - 12 = (x + 4)(x - 3)$, so $x = -4$ or $x = 3$.

4. (Core) Solve $3x^2 - 2x - 7 = 0$. Give exact answers in simplest form, and decimals to two places.

Solution

$D = (-2)^2 - 4(3)(-7) = 4 + 84 = 88$, and $\sqrt{88} = 2\sqrt{22}$.

$$x = \frac{2 \pm 2\sqrt{22}}{6} = \frac{1 \pm \sqrt{22}}{3}$$

So $x \approx 1.90$ or $x \approx -1.23$.

5. (Core) How many zeros does $f(x) = -x^2 + 4x - 5$ have? Describe what its graph looks like.

Solution

$D = 4^2 - 4(-1)(-5) = 16 - 20 = -4$. Negative, so no zeros.

The parabola opens down ($a \lt 0$) and never reaches the $x$-axis, so its vertex must be below the axis.

6. (Core) Find the zeros of $f(x) = 3(x + 2)^2 - 12$.

Solution

$$3(x + 2)^2 = 12 \quad\Rightarrow\quad (x + 2)^2 = 4 \quad\Rightarrow\quad x + 2 = \pm 2$$

So $x = 0$ or $x = -4$.

7. (Core) For which values of $k$ does $f(x) = 2x^2 - 8x + k$ have two zeros?

Solution

Two zeros means $D \gt 0$:

$$(-8)^2 - 4(2)(k) \gt 0 \quad\Rightarrow\quad 64 - 8k \gt 0 \quad\Rightarrow\quad k \lt 8$$

8. (Core) A ball is thrown upward. Its height in metres after $t$ seconds is $h(t) = -4.9t^2 + 19.6t + 1.5$. When does it hit the ground? Round to two decimal places.

Solution

Solve $h(t) = 0$ with $a = -4.9$, $b = 19.6$, $c = 1.5$:

$$D = 19.6^2 - 4(-4.9)(1.5) = 384.16 + 29.4 = 413.56$$$$t = \frac{-19.6 \pm \sqrt{413.56}}{-9.8}$$

This gives $t \approx 4.08$ or $t \approx -0.08$. Time can’t be negative, so the ball lands after about $4.08$ seconds.

9. (Challenge) For which values of $k$ does $kx^2 + 4x + 1 = 0$ have no real roots?

Solution

No real roots means $D \lt 0$:

$$4^2 - 4(k)(1) \lt 0 \quad\Rightarrow\quad 16 - 4k \lt 0 \quad\Rightarrow\quad k \gt 4$$

(Any such $k$ is positive, so the equation really is quadratic.)