Quadratic equations and solving by factorising – Quadratic Equations (Senior 3 Mathematics) | Ganzaa

Quadratic Equations

Lessons

Senior 3 · Mathematics

Quadratic Equations

5 lessons

Lesson 1 of 5

Quadratic equations and solving by factorising

24 min

From a garden to an equation

A farmer in Bugesera marks out a rectangular vegetable plot. She wants its length to be 3 m more than its width, and its area to be $40\ \mathrm{m^2}$ . If the width is $w$ metres, then

w(w + 3) = 40, \qquad\text{that is}\qquad w^2 + 3w - 40 = 0.

The unknown is squared, so the linear methods of S1 are not enough. In S2 you learned to factorise expressions such as $w^{2} + 3 w - 40$ . This lesson turns that skill into a way of solving equations, and you will find the width of the plot in Worked example 3.

Definition

Quadratic equation: an equation in one unknown that can be written in the standard form

ax^2 + bx + c = 0, \qquad a \ne 0,

where $a$ , $b$ and $c$ are real numbers (the coefficients).

Example: $2 x^{2} - 5 x - 3 = 0$ is quadratic, with $a = 2$ , $b = - 5$ , $c = - 3$ . The equations $3 x - 7 = 0$ (no $x^{2}$ term) and $x^{3} - x = 0$ (a cube) are not quadratic.

Definition

Root (or solution) of an equation: a value of the unknown that makes the equation true.

Notation: roots are often written $x_{1}$ and $x_{2}$ , or listed as a solution set, for example $S = \{-2, 3\}$ .

Example: $x = 3$ is a root of $x^{2} - x - 6 = 0$ , because $3^{2} - 3 - 6 = 9 - 3 - 6 = 0$ . A quadratic equation has at most two real roots.

First, the standard form

Before you can use any method in this unit, move every term to one side so that the other side is 0, and collect like terms.

Equation as given	Standard form	$a$	$b$	$c$
$x^{2} = 5 x$	$x^{2} - 5 x = 0$	1	$- 5$	0
$3 x^{2} = 12$	$3 x^{2} - 12 = 0$	3	0	$- 12$
$(x + 2) (x - 1) = 4$	$x^{2} + x - 6 = 0$	1	1	$- 6$
$2 x (x - 3) = x - 5$	$2 x^{2} - 7 x + 5 = 0$	2	$- 7$	5
$x (x + 4) = x^{2} + 1$	$4 x - 1 = 0$	not quadratic: the $x^{2}$ terms cancel

Important

The zero-product rule. If the product of two numbers is zero, then at least one of the numbers is zero:

p \times q = 0 \quad\Longrightarrow\quad p = 0 \ \text{ or } \ q = 0.

Why: suppose $p \ne 0$ . Then you may divide both sides of $p q = 0$ by $p$ , which gives $q = \frac{0}{p} = 0$ . So if $p$ is not zero, $q$ must be.

The rule works only for zero. From $p q = 6$ you can conclude nothing about $p$ or $q$ on their own.

Solving by factorising

Write the equation in standard form, $a x^{2} + b x + c = 0$ .
Factorise the left-hand side, using the S2 methods: a common factor, the difference of two squares, a perfect square, or two numbers with the right sum and product.
Set each factor equal to zero (the zero-product rule).
Solve each of the two linear equations.
Check each root in the original equation.

The roots have a picture. The graph of $y = x^{2} - x - 6$ is a U-shaped curve called a parabola. It crosses the $x$ -axis exactly where $y = 0$ , that is, at the roots of $x^{2} - x - 6 = 0$ .

The parabola $y = x^{2} - x - 6 = (x - 3) (x + 2)$ meets the $x$ -axis at $x = - 2$ and $x = 3$ : the roots of $x^{2} - x - 6 = 0$ .

Worked example 1: two kinds of trinomial

Solve (a) $x^{2} - x - 6 = 0$ (b) $3 x^{2} - 10 x - 8 = 0$ .

Solution

(a) Find two numbers with product $- 6$ and sum $- 1$ : they are $- 3$ and $2$ . So

(x - 3)(x + 2) = 0 \;\Rightarrow\; x - 3 = 0 \ \text{ or } \ x + 2 = 0 \;\Rightarrow\; x = 3 \ \text{ or } \ x = -2.

Check $x = - 2$ : $(- 2)^{2} - (- 2) - 6 = 4 + 2 - 6 = 0$ . ✓

(b) Here $a = 3$ , so find two numbers with product $ac = 3 \times (-8) = -24$ and sum $- 10$ : they are $- 12$ and $2$ . Split the middle term and group:

3 x^{2} - 12 x + 2 x - 8 = 3 x (x - 4) + 2 (x - 4) = (x - 4) (3 x + 2) = 0.

So $x = 4$ or $3 x + 2 = 0$ , which gives $x = -\frac{2}{3}$ . Check $x = 4$ : $3 (16) - 40 - 8 = 0$ . ✓

Definition

Repeated root (double root): a root that comes from two equal factors, so the equation has only one distinct root.

Notation: $(x - k)^{2} = 0$ has the repeated root $x = k$ .

Example: $x^{2} - 10 x + 25 = (x - 5)^{2} = 0$ has the repeated root $x = 5$ . Its parabola touches the $x$ -axis at one point instead of crossing it twice.

Worked example 2: special cases

Solve (a) $5 x^{2} = 20 x$ (b) $4 x^{2} - 25 = 0$ (c) $x^{2} - 10 x + 25 = 0$ .

Solution

(a) Standard form: $5 x^{2} - 20 x = 0$ . Common factor: $5 x (x - 4) = 0$ , so $x = 0$ or $x = 4$ .

(b) Difference of two squares: $(2 x)^{2} - 5^{2} = (2 x - 5) (2 x + 5) = 0$ , so $x = \frac{5}{2}$ or $x = -\frac{5}{2}$ .

Warning

Never divide both sides by $x$ . In (a) above, dividing $5 x^{2} = 20 x$ by $5 x$ gives only $x = 4$ and loses the root $x = 0$ : you cannot divide by $x$ when $x$ might be 0. Factorise instead.

Make the right-hand side 0 first. $x (x - 3) = 10$ does not mean $x = 10$ or $x - 3 = 10$ . Rewrite it as $x^{2} - 3 x - 10 = 0$ , then $(x - 5) (x + 2) = 0$ , so $x = 5$ or $x = - 2$ .

Worked example 3: the plot in Bugesera

Find the width and length of the plot whose length is 3 m more than its width and whose area is $40\ \mathrm{m^2}$ .

Width $w$ , length $w + 3$ , area $40\ \mathrm{m^2}$ .

Solution $w (w + 3) = 40$ , so $w^{2} + 3 w - 40 = 0$ . Two numbers with product $- 40$ and sum $3$ are $8$ and $- 5$ :

(w + 8)(w - 5) = 0 \;\Rightarrow\; w = -8 \ \text{ or } \ w = 5.

A width cannot be negative, so reject $w = - 8$ . The plot is 5 m wide and $5 + 3 = 8$ m long. Check: $5 \times 8 = 40$ . ✓

Worked example 4: working backwards from the roots

Write a quadratic equation, with whole-number coefficients, whose roots are $2$ and $- 5$ .

Solution If $x = 2$ is a root, then $x - 2$ is a factor; if $x = - 5$ is a root, then $x + 5$ is a factor. So

(x - 2)(x + 5) = 0 \quad\Longrightarrow\quad x^2 + 3x - 10 = 0.

Any multiple, such as $2 x^{2} + 6 x - 20 = 0$ , has the same roots.

Tip

Zero first, factorise second. The zero-product rule needs a 0 on one side. If you see any other number on the right, move it across before you factorise.

Khan Academy: Solving a quadratic equation by factoring (6 min). Watch how the left-hand side is factorised by splitting the middle term and grouping, and how each factor is then set equal to zero.

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Practice

Solve by factorising, and check your roots. The last three are harder.

Write in standard form and give $a$ , $b$ and $c$ : (a) $x^{2} = 7 x - 10$ (b) $(x - 3)^{2} = 2 x$ .
$x^{2} - 7 x + 10 = 0$
$x^{2} + 2 x - 15 = 0$
$3 x^{2} = 27 x$
$9 x^{2} - 16 = 0$
$2 x^{2} - x - 15 = 0$
$x (x + 1) = 12$
Write a quadratic equation with whole-number coefficients whose roots are $4$ and $-\frac{1}{3}$ .

Show answersHide answers

(a) $x^{2} - 7 x + 10 = 0$ : $a = 1$ , $b = - 7$ , $c = 10$ . (b) $x^{2} - 6 x + 9 = 2 x$ , so $x^{2} - 8 x + 9 = 0$ : $a = 1$ , $b = - 8$ , $c = 9$ .
$(x - 2) (x - 5) = 0$ : $x = 2$ or $x = 5$ .
$(x + 5) (x - 3) = 0$ : $x = - 5$ or $x = 3$ .
$3 x^{2} - 27 x = 0$ , $3 x (x - 9) = 0$ : $x = 0$ or $x = 9$ .
$(3 x - 4) (3 x + 4) = 0$ : $x = \frac{4}{3}$ or $x = -\frac{4}{3}$ .
$a c = - 30$ ; $- 6$ and $5$ : $2 x^{2} - 6 x + 5 x - 15 = (x - 3) (2 x + 5) = 0$ : $x = 3$ or $x = -\frac{5}{2}$ .
$x^{2} + x - 12 = 0$ , $(x + 4) (x - 3) = 0$ : $x = - 4$ or $x = 3$ .
$(x - 4) (3 x + 1) = 0$ , so $3 x^{2} - 11 x - 4 = 0$ .

Sources and credits

7.6: Quadratic Equations, 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.
Photo: ParabolicWaterTrajectory.jpg, GuidoB, Wikimedia Commons, licensed under CC BY-SA 3.0. Used unchanged.
Video: Solving a quadratic equation by factoring, Khan Academy, embedded from YouTube.

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

1From a garden to an equation

2First, the standard form

3Solving by factorising

From a garden to an equation

First, the standard form

Solving by factorising