Quadratic equations: factorisation, formula and completing the square

A quadratic equation can be written in the form

$ax^2+bx+c=0,$

where $a$, $b$ and $c$ are constants and $a\ne0$. It may have two distinct real solutions, one repeated real solution, or no real solutions.

Solving quadratics is a foundation for A-level algebra, coordinate geometry, trigonometry, calculus and mechanics. The essential decisions are to put the equation equal to zero, recognise its structure, and choose an efficient method.

Prerequisites

You should already be able to:

• expand and factorise quadratic expressions;
• solve linear equations;
• simplify fractions and surds;
• substitute values into expressions.

Review algebraic manipulation and factorisation, linear equations or indices, roots and surds if needed.

What does solving mean?

A solution, also called a root, is a value of $x$ that makes the equation true. For example, $x=2$ and $x=3$ solve

$x^2-5x+6=0$

because

$2^2-5(2)+6=0\qquad\text{and}\qquad3^2-5(3)+6=0.$

A quadratic can have more than one solution, so continue until every branch has been considered.

First write the equation equal to zero

Factorisation methods rely on comparing a product with zero. Rearrange all terms to one side before factorising.

Worked example 1: rearrange first

Solve $x^2+4x=12$.

Subtract $12$ from both sides:

$x^2+4x-12=0.$

Now factorise:

$$\begin{aligned} (x+6)(x-2)&=0,\\ x+6=0\quad&\text{or}\quad x-2=0,\\ x&=-6\quad\text{or}\quad x=2. \end{aligned}$$

Writing $(x+6)(x-2)=12$ would not help. A product equal to $12$ can arise in many ways.

Solving by factorisation

The zero-product property states

$uv=0\quad\Longrightarrow\quad u=0\text{ or }v=0.$

This is why a factorised quadratic can be solved by setting each factor equal to zero.

Worked example 2: a monic quadratic

Solve $x^2-7x+10=0$.

Find two numbers with product $10$ and sum $-7$. They are $-5$ and $-2$:

$$\begin{aligned} x^2-7x+10&=0,\\ (x-5)(x-2)&=0. \end{aligned}$$

Therefore

$x=5\quad\text{or}\quad x=2.$

Notice the sign reversal: $x-5=0$ gives $x=5$.

Worked example 3: a non-monic quadratic

Solve $6x^2+x-2=0$.

Since $ac=6(-2)=-12$, find two numbers with product $-12$ and sum $1$. Use $4$ and $-3$ to split the middle term:

$$\begin{aligned} 6x^2+x-2&=0,\\ 6x^2+4x-3x-2&=0,\\ 2x(3x+2)-(3x+2)&=0,\\ (2x-1)(3x+2)&=0. \end{aligned}$$

Hence

$2x-1=0\quad\text{or}\quad3x+2=0,$

so

$x=\frac12\quad\text{or}\quad x=-\frac23.$

Worked example 4: expose a common factor

Solve $5x^2=20x$.

Bring all terms to one side and factorise fully:

$$\begin{aligned} 5x^2-20x&=0,\\ 5x(x-4)&=0. \end{aligned}$$

Thus $x=0$ or $x=4$.

Misconception: cancelling the unknown

Dividing $5x^2=20x$ by $5x$ produces $x=4$, but silently loses $x=0$ because division by zero is not valid. Move everything to one side and factorise when the unknown is a common factor.

Repeated roots

Solve $x^2-6x+9=0$.

$$\begin{aligned} x^2-6x+9&=(x-3)^2,\\ (x-3)^2&=0,\\ x&=3. \end{aligned}$$

Both identical factors give the same value. This is one repeated root, not two different solutions.

Self-check 1

Solve by factorisation:

1. $x^2+x-20=0$;
2. $3x^2-11x-4=0$;
3. $2x^2+8x=0$;
4. $4x^2=25$.

Answers

1. $(x+5)(x-4)=0$, so $x=-5$ or $x=4$.
2. $(3x+1)(x-4)=0$, so $x=-\dfrac13$ or $x=4$.
3. $2x(x+4)=0$, so $x=0$ or $x=-4$.
4. $4x^2-25=(2x-5)(2x+5)=0$, so $x=\pm\dfrac52$.

Solving by the quadratic formula

For

$ax^2+bx+c=0,$

the quadratic formula is

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

It works for every quadratic equation. Use it when factorisation is not apparent or when the question requests the formula.

A reliable routine is:

1. rearrange into $ax^2+bx+c=0$;
2. identify $a$, $b$ and $c$, including their signs;
3. substitute using brackets;
4. simplify exactly before giving any decimal approximation.

Worked example 5: two irrational roots

Solve $2x^2+3x-4=0$, giving exact answers.

Here $a=2$, $b=3$ and $c=-4$. Therefore

$$\begin{aligned} x&=\frac{-3\pm\sqrt{3^2-4(2)(-4)}}{2(2)}\\ &=\frac{-3\pm\sqrt{9+32}}{4}\\ &=\frac{-3\pm\sqrt{41}}{4}. \end{aligned}$$

The exact solutions are

$\boxed{x=\frac{-3+\sqrt{41}}4\quad\text{or}\quad x=\frac{-3-\sqrt{41}}4}.$

To three significant figures, these are $x\approx0.851$ and $x\approx-2.35$.

Worked example 6: a negative coefficient

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

First rearrange with a positive $x^2$ coefficient:

$2x^2+3x-5=0.$

Now $a=2$, $b=3$ and $c=-5$:

$$\begin{aligned} x&=\frac{-3\pm\sqrt{3^2-4(2)(-5)}}{4}\\ &=\frac{-3\pm\sqrt{49}}4\\ &=\frac{-3\pm7}{4}. \end{aligned}$$

Thus $x=1$ or $x=-\dfrac52$. Although the formula works, the rearranged quadratic also factorises as $(2x+5)(x-1)$.

Misconception: the denominator covers only the square root

In

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

the entire numerator is divided by $2a$. Enter each branch into a calculator with outer brackets, for example

$\frac{-3+\sqrt{41}}4=(-3+\sqrt{41})\div4.$

Also, $-b$ means the opposite of $b$. If $b=-7$, then $-b=7$.

Exact and approximate answers

Keep a surd answer exact unless the question requests a decimal or the context requires one. Round only at the end. Premature rounding can make later calculations inaccurate.

Self-check 2

Use the quadratic formula. Give exact answers where possible.

1. $x^2+4x-1=0$;
2. $3x^2-2x-2=0$;
3. $4x^2+12x+9=0$.

Answers

1. $x=\dfrac{-4\pm\sqrt{20}}2=-2\pm\sqrt5$.
2. $x=\dfrac{2\pm\sqrt{4+24}}6=\dfrac{1\pm\sqrt7}{3}$.
3. $x=\dfrac{-12\pm\sqrt{144-144}}8=-\dfrac32$. This is a repeated root.

Solving by completing the square

Completing the square rewrites a quadratic in the form

$(x+p)^2=q,$

which can be solved by taking square roots. Remember that if $u^2=q$ and $q>0$, then

$u=\pm\sqrt q.$

Worked example 7: complete the square

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

Half the coefficient of $x$ is $3$:

$x^2+6x=(x+3)^2-9.$

Therefore

$$\begin{aligned} (x+3)^2-9-7&=0,\\ (x+3)^2&=16,\\ x+3&=\pm4,\\ x&=-3\pm4. \end{aligned}$$

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

Worked example 8: irrational answers

Solve $x^2-8x+3=0$.

$$\begin{aligned} x^2-8x+3&=0,\\ (x-4)^2-16+3&=0,\\ (x-4)^2&=13,\\ x-4&=\pm\sqrt{13},\\ x&=4\pm\sqrt{13}. \end{aligned}$$

Misconception: forgetting both square roots

From $(x-4)^2=13$, writing only $x-4=\sqrt{13}$ loses a solution. Both $\sqrt{13}$ and $-\sqrt{13}$ square to $13$.

The method and its connection with turning points are developed further in completing the square.

When there are no real solutions

Consider

$x^2+2x+5=0.$

Completing the square gives

$(x+1)^2+4=0,$

so

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

No real number has a negative square. Therefore the equation has no real solutions. At this stage, do not treat a calculator error from $\sqrt{-4}$ as a numerical accident. The full A-level lesson uses the discriminant to classify roots efficiently.

Choosing a method

Use the form of the equation to guide you:

• Factorise when integer or simple rational factors are visible. This is usually quickest.
• Use the quadratic formula when factorisation is unclear. It is the dependable general method.
• Complete the square when requested, when the expression is close to a square, or when the completed form will be useful later.
• Use square roots directly for equations such as $(x-2)^2=11$ or $x^2=7$.

For example,

$3(x-1)^2=15$

does not need expansion:

$$\begin{aligned} (x-1)^2&=5,\\ x-1&=\pm\sqrt5,\\ x&=1\pm\sqrt5. \end{aligned}$$

Forming and interpreting a quadratic

Worked example 9: dimensions of a rectangle

A rectangle is $3$ cm longer than it is wide and has area $54\text{ cm}^2$. Find its dimensions.

Let the width be $x$ cm, so the length is $(x+3)$ cm. Then

$x(x+3)=54.$

Rearrange and solve:

$$\begin{aligned} x^2+3x-54&=0,\\ (x+9)(x-6)&=0. \end{aligned}$$

Thus $x=-9$ or $x=6$. A physical length must be positive, so reject $x=-9$ in this context. The rectangle is $6$ cm by $9$ cm.

The negative value is still an algebraic root. It is rejected because it does not satisfy the model’s domain.

Worked example 10: consecutive integers

The product of two consecutive positive integers is $132$. Find the integers.

Let the smaller integer be $n$. Then

$n(n+1)=132,$

so

$$\begin{aligned} n^2+n-132&=0,\\ (n+12)(n-11)&=0. \end{aligned}$$

Therefore $n=-12$ or $n=11$. The condition says positive integers, so $n=11$. The integers are $11$ and $12$.

Checking solutions

Substitute each answer into the original equation, especially after rearranging or modelling.

For $2x^2+3x-5=0$, check $x=-\dfrac52$:

$$2\left(-\frac52\right)^2+3\left(-\frac52\right)-5 =\frac{25}{2}-\frac{15}{2}-5 =0.$$

A quick numerical check can reveal a sign or transcription error. It does not replace a valid derivation.

Mixed self-check

1. Solve $x^2=5x$.
2. Solve $2x^2-7x+3=0$ by factorisation.
3. Solve $x^2-2x-6=0$, giving exact answers.
4. Solve $(2x+1)^2=12$, giving exact answers.
5. A rectangle has width $x$ metres, length $(x+5)$ metres and area $24\text{ m}^2$. Find its dimensions.
6. Explain why $x^2-4x+8=0$ has no real solutions.

Answers

1. $x^2-5x=x(x-5)=0$, so $x=0$ or $x=5$.
2. $(2x-1)(x-3)=0$, so $x=\dfrac12$ or $x=3$.
3. By the formula, $x=\dfrac{2\pm\sqrt{4+24}}2=1\pm\sqrt7$.
4. $2x+1=\pm\sqrt{12}=\pm2\sqrt3$, so $x=\dfrac{-1\pm2\sqrt3}{2}$.
5. $x(x+5)=24$ gives $(x+8)(x-3)=0$. Reject $x=-8$, so the dimensions are $3$ m by $8$ m.
6. $x^2-4x+8=(x-2)^2+4$. Setting this equal to zero would require $(x-2)^2=-4$, which is impossible for real $x$.

Key points

• Rearrange a quadratic into $ax^2+bx+c=0$ before choosing a method.
• Factorisation works because a product equal to zero must have at least one zero factor.
• Never divide by an expression containing the unknown unless you have separately considered when that expression is zero.
• In the quadratic formula, copy signs carefully and divide the whole numerator by $2a$.
• Taking a square root normally creates two branches, shown by $\pm$.
• Keep exact surd answers until a decimal approximation is requested.
• Check answers in the original equation and reject roots only when the context makes them inadmissible.

Next steps

Continue to quadratic functions and equations for quadratic graphs, the discriminant, equations quadratic in another expression, parameters and modelling. Then study completing the square and linear and quadratic inequalities.