Quadratic equations: Grade 12 Mathematics

By the end of this lesson you should be able to

Solve quadratic equations by factorising
Solve quadratic equations using the quadratic formula and by completing the square
Use the discriminant to describe the roots of a quadratic equation
Form and solve quadratic equations from word problems

What is a quadratic equation?

A quadratic equation is an equation that can be written in the standard form

ax^2 + bx + c = 0 \qquad (a \neq 0)

where $a$ , $b$ and $c$ are numbers. The highest power of $x$ is $2$ , and a quadratic equation usually has two solutions, also called roots. Some have one repeated root, and some have no real roots at all.

Always begin by moving every term to one side so that the other side is $0$ . Beginning with an equation that is not in standard form is the most common cause of lost marks.

Method 1: factorising

If the quadratic factorises, this is the quickest method. It relies on one fact: if two numbers multiply to give $0$ , at least one of them must be $0$ .

Example 1. Solve $x^2 + 5x + 6 = 0$ .

Find two numbers that multiply to $6$ and add to $5$ . They are $2$ and $3$ .

(x + 2)(x + 3) = 0

So $x + 2 = 0$ or $x + 3 = 0$ , which gives $x = -2$ or $x = -3$ .

Example 2. Solve $2x^2 + 7x + 3 = 0$ .

Here $a \neq 1$ , so the numbers must multiply to $a \times c = 6$ and add to $7$ . They are $1$ and $6$ . Split the middle term:

2x^2 + x + 6x + 3 = 0

x(2x + 1) + 3(2x + 1) = 0

(2x + 1)(x + 3) = 0

So $x = -\dfrac{1}{2}$ or $x = -3$ .

Method 2: the quadratic formula

When a quadratic does not factorise neatly, use the formula.

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

Write down $a$ , $b$ and $c$ first, keeping their signs, then substitute carefully using brackets.

Example 3. Solve $3x^2 - 5x - 2 = 0$ .

Here $a = 3$ , $b = -5$ and $c = -2$ .

x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-2)}}{2(3)} = \frac{5 \pm \sqrt{25 + 24}}{6} = \frac{5 \pm 7}{6}

So $x = \dfrac{12}{6} = 2$ or $x = \dfrac{-2}{6} = -\dfrac{1}{3}$ .

Example 4. Solve $x^2 - 4x - 3 = 0$ , giving your answers correct to 2 decimal places.

Here $a = 1$ , $b = -4$ and $c = -3$ .

x = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2}

$x = \dfrac{4 + 5.2915}{2} = 4.65$ or $x = \dfrac{4 - 5.2915}{2} = -0.65$ .

Method 3: completing the square

Completing the square rewrites the equation as $(x + p)^2 = q$ , which you can solve by taking square roots. It is also the method used to derive the formula.

Example 5. Solve $x^2 + 6x - 7 = 0$ by completing the square.

Move the constant across: $x^2 + 6x = 7$ .

Add the square of half the coefficient of $x$ , which is $3^2 = 9$ , to both sides:

x^2 + 6x + 9 = 16 \quad\Rightarrow\quad (x + 3)^2 = 16

Take square roots, remembering both signs: $x + 3 = \pm 4$ .

So $x = 1$ or $x = -7$ .

The discriminant

The part of the formula under the square root, $b^2 - 4ac$ , is called the discriminant. It tells you about the roots without solving the equation.

Discriminant	Number of real roots
$b^2 - 4ac > 0$	Two different real roots
$b^2 - 4ac = 0$	One repeated (equal) real root
$b^2 - 4ac < 0$	No real roots

Example 6. Find the values of $k$ for which $x^2 + kx + 9 = 0$ has equal roots.

Equal roots need $b^2 - 4ac = 0$ :

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

Quadratic equations in word problems

Define a letter, form the equation, solve it, then decide whether each root makes sense. A length cannot be negative.

Example 7. The length of a rectangle is 3 m more than its width. Its area is 40 m². Find the dimensions.

Let the width be $w$ metres, so the length is $w + 3$ .

w(w + 3) = 40 \quad\Rightarrow\quad w^2 + 3w - 40 = 0 \quad\Rightarrow\quad (w + 8)(w - 5) = 0

So $w = -8$ or $w = 5$ . A width cannot be negative, so $w = 5$ .

The rectangle is 5 m wide and 8 m long. Check: $5 \times 8 = 40$ . ✓

Common mistakes

Not writing the equation in standard form first. For $x^2 = 5x$ , do not divide by $x$ , because that loses the solution $x = 0$ . Rewrite it as $x^2 - 5x = 0$ , so $x(x - 5) = 0$ and $x = 0$ or $x = 5$ .

Sign errors in the formula. If $b = -5$ , then $-b = 5$ and $b^2 = 25$ . Use brackets when substituting negative numbers.

Forgetting the $\pm$ . Square roots give two values. Leaving out one of them gives half the answer.

Accepting every root in a word problem. Always test each root against the situation. A negative length, age or number of people is rejected.

Practice questions

Solve $x^2 - 7x + 12 = 0$ .
Solve $x^2 - 9 = 0$ .
Solve $2x^2 - x - 6 = 0$ .
Solve $x^2 + 2x - 5 = 0$ , giving your answers correct to 2 decimal places.
Solve $x^2 - 8x + 3 = 0$ by completing the square, giving your answers correct to 2 decimal places.
Find the values of $k$ for which $x^2 + kx + 16 = 0$ has equal roots.
Without solving, state the nature of the roots of $2x^2 - 3x + 5 = 0$ .
The product of two consecutive positive integers is 132. Find the integers.

Show answers

$(x - 3)(x - 4) = 0$ , so $x = 3$ or $x = 4$
$x^2 = 9$ , so $x = 3$ or $x = -3$
$(2x + 3)(x - 2) = 0$ , so $x = -\dfrac{3}{2}$ or $x = 2$
$x = \dfrac{-2 \pm \sqrt{24}}{2} = -1 \pm \sqrt{6}$ , so $x = 1.45$ or $x = -3.45$
$(x - 4)^2 = 13$ , so $x = 4 \pm \sqrt{13}$ , giving $x = 7.61$ or $x = 0.39$
$k^2 - 64 = 0$ , so $k = 8$ or $k = -8$
$b^2 - 4ac = 9 - 40 = -31$ , which is less than $0$ , so there are no real roots
Let the integers be $n$ and $n + 1$ . Then $n^2 + n - 132 = 0$ , so $(n + 12)(n - 11) = 0$ . Since $n$ is positive, $n = 11$ . The integers are 11 and 12.

Summary

Rearrange to $ax^2 + bx + c = 0$ before doing anything else.
Factorise when you can, use the formula when you cannot, and use completing the square when the question asks for it.
The discriminant $b^2 - 4ac$ tells you whether there are two, one or no real roots.
In word problems, check each root against the situation before giving your answer.

Now practise this topic

Real questions show you what still needs work. Try a past paper under timed conditions, or take a short quiz.

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