2.4 Solving quadratic equations, including in a function of the unknown — Learn

Prerequisites: factorising from 2.3; the discriminant; completing the square; the index laws from 2.1 and surds from 2.2.

By the end of this section you can: solve a quadratic equation by factorising, by completing the square, and by the formula; choose the method that fits; and recognise and solve an equation that is quadratic in a function of the unknown, such as $x^4$ , $2^x$ or $\sqrt{x}$ .

1. Choose a method

A quadratic equation has the form $ax^2 + bx + c = 0$ . Four methods solve it, and the numbers tell you which to use.

Method	Use it when	What it gives
Factorising	the roots are whole numbers or simple fractions	exact roots, quickly
Completing the square	the coefficient $a$ is $1$ and the roots contain a surd	exact roots, and the vertex
The formula	factorising fails and $a \neq 1$	exact roots, always
A graph	the question asks where the curve meets a line	an estimate

The formula solves every quadratic equation:

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

The expression $b^2 - 4ac$ is the discriminant from 2.3. When it is negative, there is no real solution and the method stops there.

2. Factorising and the formula

Example 1. Solve $2x^2 + 5x - 3 = 0$ .

Look for two numbers that multiply to $2 \times (-3) = -6$ and add to $5$ . The numbers are $6$ and $-1$ . Split the middle term and factorise in pairs:

2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3).

A product is zero when one of its factors is zero:

2x - 1 = 0 \quad \text{or} \quad x + 3 = 0,

so $x = \frac{1}{2}$ or $x = -3$ .

Example 2. Solve $x^2 - 3x - 2 = 0$ , giving the answer in surd form.

No whole-number pair multiplies to $-2$ and adds to $-3$ , so use the formula with $a = 1$ , $b = -3$ and $c = -2$ :

x = \frac{3 \pm \sqrt{(-3)^2 - 4 \times 1 \times (-2)}}{2}.

Work out the discriminant:

x = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}.

The two roots are $\frac{3 - \sqrt{17}}{2}$ and $\frac{3 + \sqrt{17}}{2}$ .

3. Completing the square to solve

When $a = 1$ and the roots contain a surd, completing the square is often quicker than the formula.

Example 3. Solve $x^2 + 4x - 1 = 0$ by completing the square.

Move the constant to the right:

x^2 + 4x = 1.

Complete the square on the left. Half of $4$ is $2$ , and $(x + 2)^2 = x^2 + 4x + 4$ , so

(x + 2)^2 - 4 = 1.

Add $4$ to both sides, then take the square root:

(x + 2)^2 = 5, \qquad x + 2 = \pm \sqrt{5}.

x = -2 \pm \sqrt{5}.

4. Quadratic in a function of the unknown

An equation can be a quadratic even when it is not written with an $x^2$ . Look for a repeated expression, such as $x^2$ inside an $x^4$ , or $2^x$ inside a $2^{2x}$ . Replace the repeated expression with a single letter, solve the quadratic, then put the expression back.

Example 4. Solve $x^4 - 5x^2 + 4 = 0$ .

The term $x^4$ is $\left(x^2\right)^2$ , so let $t = x^2$ . Then

t^2 - 5t + 4 = 0.

Factorise:

(t - 1)(t - 4) = 0,

so $t = 1$ or $t = 4$ . Now replace $t$ with $x^2$ :

x^2 = 1 \quad \text{or} \quad x^2 = 4.

Taking both signs of each square root gives $x = \pm 1$ and $x = \pm 2$ .

Example 5. Solve $2^{2x} - 5 \times 2^x + 4 = 0$ .

The term $2^{2x}$ is $\left(2^x\right)^2$ , so let $t = 2^x$ . Then

t^2 - 5t + 4 = 0,

which factorises as $(t - 1)(t - 4) = 0$ . So $t = 1$ or $t = 4$ . Replace $t$ with $2^x$ :

2^x = 1 \quad \text{or} \quad 2^x = 4.

Since $2^0 = 1$ and $2^2 = 4$ , the solutions are $x = 0$ and $x = 2$ .

Example 6. Solve $1 - \frac{3}{x} + \frac{2}{x^2} = 0$ , where $x \neq 0$ .

The expression $\frac{1}{x}$ appears twice, once squared, so let $t = \frac{1}{x}$ :

2t^2 - 3t + 1 = 0.

Factorise as $(2t - 1)(t - 1) = 0$ , so $t = \frac{1}{2}$ or $t = 1$ . Replace $t$ with $\frac{1}{x}$ :

\frac{1}{x} = \frac{1}{2} \quad \text{or} \quad \frac{1}{x} = 1,

so $x = 2$ or $x = 1$ . Both values are allowed, because neither is zero.

Watch out. The substitution is not finished until you replace $t$ and check every value.

Stopping at $t$ . The answer to Example 4 is a value of $x$ , not a value of $t$ . Writing $t = 1$ or $t = 4$ leaves the question unanswered.

Throwing away a root. $x^2 = 4$ has two solutions, $x = 2$ and $x = -2$ . Do not drop the negative one.

Adding a root by squaring. Squaring both sides of $\sqrt{x + 5} = x - 1$ gives $x^2 - 3x - 4 = 0$ , so $x = 4$ or $x = -1$ . The value $x = -1$ is extraneous: it makes the right side $-2$ , while a square root is never negative. Re-check: substitute every answer into the original equation, and reject any value that fails.

Progress check

Return to the chapter problem. The arch is $y = 4 - \frac{x^2}{4}$ , and the lorry roof is $3.2$ m above the road.

The lorry fits under the arch at a distance $x$ from the centre if $4 - \frac{x^2}{4} \geq 3.2$ . Solve the equation $4 - \frac{x^2}{4} = 3.2$ .
The two solutions are the points where the roof grazes the arch. How wide is the arch at the height of the lorry roof?
The lorry is $2.5$ m wide. Explain why this calculation answers a different question from the one about the lorry's width.
Solve $2^{2x} - 6 \times 2^x + 8 = 0$ .

You should now be able to

choose between factorising, completing the square and the formula;
solve a quadratic equation with surd roots by completing the square;
recognise a repeated expression and substitute $t$ for it;
solve a quadratic in $t$ and then recover the values of $x$ ;
reject an extraneous root created by squaring both sides.