The quadratic formula and the discriminant – Quadratic Equations (Senior 3 Mathematics) | Ganzaa

Quadratic Equations

Lessons

Senior 3 · Mathematics

Quadratic Equations

5 lessons

Lesson 3 of 5

The quadratic formula and the discriminant

25 min

One method for every quadratic

Completing the square always works, but it takes several lines, and the arithmetic is different each time. Mathematicians therefore completed the square once, for the general equation $a x^{2} + b x + c = 0$ . The result is a formula into which you put $a$ , $b$ and $c$ and read off the roots. Part of that formula, the discriminant, even tells you how many roots there are before you solve.

Deriving the formula

Start from $a x^{2} + b x + c = 0$ with $a \ne 0$ , and follow exactly the steps of the last lesson.

\begin{aligned} x^2 + \frac{b}{a}x + \frac{c}{a} &= 0 && \text{divide by } a \\ x^2 + \frac{b}{a}x &= -\frac{c}{a} && \text{move the constant} \\ x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} &= \frac{b^2}{4a^2} - \frac{c}{a} && \text{add } \left(\frac{b}{2a}\right)^2 \\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2 - 4ac}{4a^2} && \text{common denominator} \\ x + \frac{b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a} && \text{square root property} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \end{aligned}

Important

The quadratic formula. The roots of $a x^{2} + b x + c = 0$ , where $a \ne 0$ and $b^2 - 4ac \ge 0$ , are

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

Why: it is the result of completing the square on the general equation, as shown above. The whole of $-b \pm \sqrt{b^2 - 4ac}$ is divided by $2 a$ .

Worked example 1: a formula check on an easy equation

Solve $2 x^{2} + 5 x - 3 = 0$ with the formula.

Solution $a = 2$ , $b = 5$ , $c = - 3$ . First $b^{2} - 4 a c = 25 - 4 (2) (- 3) = 25 + 24 = 49$ . Then

x = \frac{-5 \pm \sqrt{49}}{2 \times 2} = \frac{-5 \pm 7}{4}, \quad\text{so}\quad x = \frac{2}{4} = \frac{1}{2} \ \text{ or } \ x = \frac{-12}{4} = -3.

Because 49 is a perfect square, the equation also factorises: $(2 x - 1) (x + 3) = 0$ .

Worked example 2: surd roots

Solve $3 x^{2} - 4 x - 2 = 0$ . Give exact roots in simplest surd form, then to 2 decimal places.

Solution $a = 3$ , $b = - 4$ , $c = - 2$ . $b^{2} - 4 a c = 16 + 24 = 40$ , and $\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$ .

x = \frac{4 \pm 2\sqrt{10}}{6} = \frac{2 \pm \sqrt{10}}{3}.

With $\sqrt{10} \approx 3.162$ : $x \approx 1.72$ or $x \approx -0.39$ .

Worked example 3: rearrange first

Solve $5 x = 2 x^{2} - 1$ to 2 decimal places.

Solution Standard form: $2 x^{2} - 5 x - 1 = 0$ , so $a = 2$ , $b = - 5$ , $c = - 1$ . Then $- b = 5$ and $b^{2} - 4 a c = 25 + 8 = 33$ :

x = \frac{5 \pm \sqrt{33}}{4} \approx \frac{5 \pm 5.745}{4}, \quad\text{so}\quad x \approx 2.69 \ \text{ or } \ x \approx -0.19.

Definition

Discriminant of $a x^{2} + b x + c = 0$ : the number $b^{2} - 4 a c$ , the expression under the square root in the quadratic formula.

Notation: $\Delta = b^2 - 4ac$ ( $\Delta$ is the Greek capital letter delta).

Example: for $x^{2} - 2 x - 3 = 0$ , $\Delta = (-2)^2 - 4(1)(-3) = 4 + 12 = 16$ .

What the discriminant tells you

$\Delta$	Roots	Why	Parabola $y = a x^{2} + b x + c$
$\Delta \gt 0$	two different real roots	$\pm\sqrt{\Delta}$ gives two different values	crosses the $x$ -axis twice
$\Delta = 0$	one repeated root, $x = -\frac{b}{2a}$	$\pm\sqrt{0}$ adds and subtracts nothing	touches the $x$ -axis once
$\Delta \lt 0$	no real roots	a negative number has no real square root	does not meet the $x$ -axis

$y = x^{2} - 2 x - 3$ , $y = x^{2} - 2 x + 1$ and $y = x^{2} - 2 x + 3$ : only $c$ changes, but the number of roots goes from two to one to none.

Worked example 4: predicting the number of roots

Without solving, state how many real roots each equation has: (a) $3 x^{2} - 5 x + 4 = 0$ (b) $4 x^{2} - 12 x + 9 = 0$ (c) $x^{2} + x - 1 = 0$ .

Solution

(a) $\Delta = 25 - 48 = -23 \lt 0$ : no real roots.

(b) $\Delta = 144 - 144 = 0$ : one repeated root, $x = \frac{12}{8} = \frac{3}{2}$ . (Indeed $4 x^{2} - 12 x + 9 = (2 x - 3)^{2}$ .)

Worked example 5: an unknown coefficient

(a) Find the values of $k$ for which $x^{2} + k x + 9 = 0$ has equal roots. (b) Find the values of $m$ for which $x^{2} - 4 x + m = 0$ has two different real roots.

Solution

(a) Equal roots need $\Delta = 0$ : $k^{2} - 4 (1) (9) = 0$ , so $k^{2} = 36$ and $k = 6$ or $k = - 6$ . (Check: $x^{2} + 6 x + 9 = (x + 3)^{2}$ .)

(b) Two different roots need $\Delta \gt 0$ : $16 - 4m \gt 0$ , so $m \lt 4$ .

Choosing a method

The equation looks like	Best method	Example
no $x$ term, or already a square	square root property	$(x - 1)^{2} = 6$
factorises easily ( $\Delta$ is a perfect square)	factorising	$x^{2} - 5 x + 6 = 0$
does not factorise	quadratic formula	$3 x^{2} - 4 x - 2 = 0$
the question asks for it, or you need the vertex	completing the square	$x^{2} + 6 x - 3 = 0$

Warning

Signs and fraction bars. If $b = - 4$ , then $- b = + 4$ and $b^{2} = + 16$ . If $c$ is negative, $- 4 a c$ is added. And the fraction bar runs under the whole of $-b \pm \sqrt{\Delta}$ : $\frac{-b \pm \sqrt{\Delta}}{2a}$ is not $-b \pm \frac{\sqrt{\Delta}}{2a}$ .

Tip

Work out $\Delta$ first, on its own line. It catches sign mistakes early, tells you how many roots to expect, and if it is a perfect square (1, 4, 9, 16, 25, …), the equation could also have been factorised.

Corbettmaths: The Discriminant (14 min). Watch how the value of b² − 4ac is linked to the number of times the graph meets the x-axis, and how exam questions with an unknown coefficient are set up as an equation or inequality.

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Practice

Use the quadratic formula. Give exact roots and, where needed, 2 d.p. The last three are harder.

$x^{2} + 3 x - 10 = 0$
$2 x^{2} - 3 x - 4 = 0$
$x^{2} - 6 x + 4 = 0$
$5 x^{2} + 2 x - 1 = 0$
Find the discriminant and the number of real roots: (a) $x^{2} + 4 x + 5 = 0$ (b) $9 x^{2} - 6 x + 1 = 0$ (c) $2 x^{2} - 7 x + 3 = 0$ .
Find $k$ if $k x^{2} - 8 x + 2 = 0$ has equal roots.
For which values of $m$ does $x^{2} + 2 x + m = 0$ have no real roots?
Show that $x^{2} + p x - 4 = 0$ has two different real roots for every value of $p$ .

Show answersHide answers

$\Delta = 9 + 40 = 49$ : $x = \frac{-3 \pm 7}{2}$ , so $x = 2$ or $x = - 5$ .
$\Delta = 9 + 32 = 41$ : $x = \frac{3 \pm \sqrt{41}}{4} \approx 2.35$ or $- 0.85$ .
$\Delta = 36 - 16 = 20$ : $x = \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5} \approx 5.24$ or $0.76$ .
$\Delta = 4 + 20 = 24$ : $x = \frac{-2 \pm 2\sqrt{6}}{10} = \frac{-1 \pm \sqrt{6}}{5} \approx 0.29$ or $- 0.69$ .
(a) $16 - 20 = - 4$ : none (b) $36 - 36 = 0$ : one repeated root (c) $49 - 24 = 25$ : two.
$64 - 8 k = 0$ , so $k = 8$ .
$4 - 4m \lt 0$ , so $m \gt 1$ .
$\Delta = p^2 + 16$ . Since $p^2 \ge 0$ , $\Delta \ge 16 \gt 0$ for every $p$ .

Sources and credits

10.3: Solve Quadratic Equations Using the Quadratic Formula, from Elementary Algebra by Lynn Marecek, MaryAnne Anthony-Smith and Andrea Honeycutt Mathis, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 30 September 2026). Adapted by Ganzaa.
The three-parabola diagram and the table for choosing a method. Written by Ganzaa.
Video: The Discriminant, Corbettmaths, embedded from YouTube.

The Rwandan worked examples, the diagrams, the practice items and the quiz questions were written by Ganzaa.

1One method for every quadratic

2Deriving the formula

3What the discriminant tells you

4Choosing a method

One method for every quadratic

Deriving the formula

What the discriminant tells you

Choosing a method