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.

If you haven’t seen it yet, $f(x)$ just means “the $y$-value when the input is $x$”: writing $f(x) = x^2 - 4$ is the same as $y = x^2 - 4$. (You’ll meet function notation properly in Grade 11 — see function notation.)

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.

On the SAT

In the built-in Desmos calculator, type the quadratic (like y = 2x^2 + 3x - 4) and click the parabola: the grey dots on the $x$-axis are the zeros, and counting them tells you whether there are 2, 1, or 0 real solutions. Desmos shows rounded decimals (here about $0.851$ and $-2.351$), so if the choices are exact, like $\dfrac{-3 \pm \sqrt{41}}{4}$, convert them to decimals to compare or use the quadratic formula. When a parabola only just touches the axis, it’s hard to tell “one zero” from “none” by eye, so confirm with the discriminant. See using Desmos on the SAT.

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.)