International Baccalaureate · MYP 4

Quadratic Equations

⌚ ~170 min ▢ 19 blocks ✓ 79 exercises ➣ Prerequisites : Factorisation, Linear Equations, Inequalities and Formulae, Square and Cube Roots, Radicals and Surds

I Quadratic Equations

I.1 What Is a Quadratic Equation?

Definition — Quadratic Equation

A quadratic equation is an equation that can be written in the standard form: $$ax^2 + bx + c = 0$$ where $a, b,$ and $c$ are known coefficients, and $a \neq 0$. The condition $a \neq 0$ is what makes the equation quadratic.
A solution or root of the equation is a value of $x$ that makes the statement true.

Example

Consider the equation $3x^2 + 5x + 4 = 0$. Is it a quadratic equation? If yes, identify the coefficients $a$, $b$, and $c$.

Answer

Yes. It is in the form $ax^2+bx+c=0$ with $a=3$, $b=5$, $c=4$, and $a\neq 0$, so it is a quadratic equation.

Example

Are $1$ and $3$ roots of the equation $x^2 - 3x + 2 = 0$?

Answer

To check if $1$ and $3$ are roots, substitute each into the equation:

For $x=1$, $1^2 - 3 \cdot 1 + 2 = 1 - 3 + 2 = 0$.
So $1$ is a root.
For $x=3$, $3^2 - 3 \cdot 3 + 2 = 9 - 9 + 2 = 2 \neq 0$.
So $3$ is not a root.

Skills to practice

Recognising Quadratic Equations
Verifying Roots of Quadratic Equations
Identifying the Coefficients

II Solving by Factorisation

II.1 The Null Factor Law

Factorising an expression turns it into a product. This skill is essential for solving new kinds of equations, especially when an equation can be written as a product of factors equal to zero. The power of factorisation comes from a simple but fundamental property of the number zero.

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Proposition — Null Factor Law

If the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero. If $\textcolor{colordef}{A}\,\textcolor{colorprop}{B} = 0$ then $\textcolor{colordef}{A=0}$ or $\textcolor{colorprop}{B=0}$. Note: one or both of the factors can be zero. Conversely, if a factor is zero, the product is zero — so each candidate found this way is indeed a solution.

Method — Solving by Factorisation

The Null Factor Law gives a powerful strategy for solving equations of the form “product of factors = 0”:

Rearrange the equation so that all the terms are on one side and zero is on the other side.
Factorise completely the side that contains the variable(s).
Set each factor equal to zero and solve the resulting linear equations.

Example

Solve for $x$: $x(x+1)=0$

Answer

The expression is already factorised and equal to zero, so we can apply the Null Factor Law directly by setting each factor equal to zero: $$ \begin{aligned} \textcolor{colordef}{x}\,\textcolor{colorprop}{(x+1)} &= 0 \\ \textcolor{colordef}{x=0} &\text{ or }\textcolor{colorprop}{(x+1)=0} &&\text{(Null Factor Law)}\\ \textcolor{colordef}{x=0} &\text{ or }\textcolor{colorprop}{x=-1}\,. \end{aligned} $$ The equation has two solutions: $x=0$ and $x=-1$.

Analysis and Synthesis

Solving an equation is a two-step reasoning.

The analysis: if a number $x$ is a solution, then the rules force conditions on it — step by step, they narrow the search down to candidates. For $(x-1)(x+2) = 0$, the Null Factor Law leaves two candidates: $1$ and $-2$.
The synthesis: conversely, each candidate must be tested in the original equation, because the analysis only shows what a solution would have to look like.

For $x = 1$: $(1-1)(1+2) = 0 \times 3 = 0$. For $x = -2$: $(-2-1)(-2+2) = (-3) \times 0 = 0$. Both candidates pass the test: the solutions are $1$ and $-2$. The check is not a luxury — it is the second half of the reasoning.

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Skills to practice

Finding the Solutions of a Product Equal to Zero
Solving with the Null Factor Law: Level 1
Solving with the Null Factor Law: Level 2
Writing the Solution of a Product Equal to Zero

II.2 Special Forms

Before learning a general method, we first master solving equations that can be factorised using familiar patterns.

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Proposition — Common Factor ($c\equal 0$)

For equations of the form $ax^2+bx=0$, the common factor is $x$: $$x(ax+b)=0 \Leftrightarrow x=0 \text{ or } ax+b=0$$

Example

Find the roots of $x^2 - 2x = 0$.

Answer

$$\begin{aligned}x^2 - 2x &= 0 \\ x(x-2) &= 0 &&\color{gray}\text{(Factor out the common factor }x\color{gray}\text{)} \\ x = 0 &\quad \text{or} \quad x-2=0 &&\color{gray}\text{(Null factor law)} \\ x = 0 &\quad \text{or} \quad x=2 \end{aligned}$$

Proposition — Solutions of $x^2 \equal a$

Let $a$ be a number. The solutions of the equation $x^2 = a$ depend on the sign of $a$:

if $a > 0$, the equation has two solutions: $x = \sqrt{a}$ and $x = -\sqrt{a}$;
if $a = 0$, the equation has exactly one solution: $x = 0$;
if $a < 0$, the equation has no solution: the square of a number is never negative.

Proposition — Difference of Squares ($b\equal 0$)

For equations of the form $x^2-k=0$ (where $k>0$): $$x^2 - (\sqrt{k})^2 = 0 \Leftrightarrow (x - \sqrt{k})(x + \sqrt{k})=0$$

Example

Solve $x^2 - 9 = 0$.

Answer

$$\begin{aligned}x^2 - 9 &= 0 \\ x^2 - 3^2 &= 0 \\ (x - 3)(x + 3) &= 0 &&\color{gray}\text{(Difference of squares)} \\ x - 3 = 0 &\quad \text{or} \quad x + 3 = 0 &&\color{gray}\text{(Null factor law)} \\ x = 3 &\quad \text{or} \quad x = -3 \end{aligned}$$

Proposition — Perfect Squares

For equations of the form $x^2 \pm 2kx + k^2=0$: $$ \begin{aligned} x^2 + 2kx + k^2 &= (x + k)^2 = 0 \Leftrightarrow x + k = 0,\\ x^2 - 2kx + k^2 &= (x - k)^2 = 0 \Leftrightarrow x - k = 0. \end{aligned} $$

Example

Solve $x^2 + 2x + 1 = 0$.

Answer

$$\begin{aligned}x^2 + 2x + 1 &= 0 \\ (x + 1)^2 &= 0 &&\color{gray}\text{(Perfect square factorisation)} \\ x + 1 &= 0 &&\color{gray}\text{(Null factor law)} \\ x &= -1 \end{aligned}$$ The equation has one solution, $-1$, counted twice (a double root): the factor $x + 1$ appears twice in $(x + 1)^2 = (x + 1)(x + 1)$.

Skills to practice

Finding the Solutions of $ax^2 + bx \equal 0$
Finding the Solutions of a Difference of Squares
Solving Equations of the Form $(x + p)^2 \equal q$
Solving Equations with a Common Factor
Solving Equations with a Difference of Squares

II.3 Solving a Quadratic Equation by Factorising

Method — Solving by Factorisation

The primary strategy for solving quadratic equations is to use the Null Factor Law.

Write the equation in standard form, $ax^2+bx+c=0$.
Factorise the quadratic expression completely.
Apply the Null Factor Law: set each factor equal to zero.
Solve each resulting linear equation.

Example

Solve $(x - 1)(x + 2) = 0$.

Answer

$$\begin{aligned}(x-1)(x+2) &= 0\\ x-1 = 0 &\quad \text{or} \quad x+2=0 &&\color{gray}\text{(Null factor law)} \\ x = 1 &\quad \text{or} \quad x = -2 &&\color{gray}\text{(Solve each equation)}\\ \end{aligned}$$

Example

Solve each equation by factorising.

$x^2 + 5x + 6 = 0$
$x^2 = 3x + 10$
$2x^2 - 2x - 12 = 0$

Answer

The equation is already in standard form. By the Proposition “Sum and Product” ($p + q = b$ and $pq = c$ give $x^2 + bx + c = (x + p)(x + q)$), look for two numbers with sum $5$ and product $6$: they are $2$ and $3$, so $$\begin{aligned}[t] x^2 + 5x + 6 &= 0\\ (x + 2)(x + 3) &= 0 && \text{(Sum and Product)}\\ x + 2 = 0 \ &\text{ or } \ x + 3 = 0 && \text{(Null Factor Law)}\\ x = -2 \ &\text{ or } \ x = -3. \end{aligned}$$ The solutions are the opposites of the two numbers found: the factor $x + 2$ is zero for $x = -2$, not for $x = 2$. Check: $(-2)^2 + 5 \times (-2) + 6 = 4 - 10 + 6 = 0$ and $(-3)^2 + 5 \times (-3) + 6 = 9 - 15 + 6 = 0$. The solutions are $-3$ and $-2$.
First bring every term to the left side, to get the standard form $x^2 - 3x - 10 = 0$; the two numbers with sum $-3$ and product $-10$ are $-5$ and $2$, so $$\begin{aligned}[t] x^2 &= 3x + 10\\ x^2 - 3x - 10 &= 0 && \text{(subtract } 3x + 10\text{)}\\ (x - 5)(x + 2) &= 0 && \text{(Sum and Product)}\\ x - 5 = 0 \ &\text{ or } \ x + 2 = 0 && \text{(Null Factor Law)}\\ x = 5 \ &\text{ or } \ x = -2. \end{aligned}$$ Check in the original equation: for $x = 5$, $5^2 = 25$ and $3 \times 5 + 10 = 25$; for $x = -2$, $(-2)^2 = 4$ and $3 \times (-2) + 10 = 4$. The solutions are $-2$ and $5$.
First take out the common factor $2$; the two numbers with sum $-1$ and product $-6$ are $-3$ and $2$, so $$\begin{aligned}[t] 2x^2 - 2x - 12 &= 0\\ 2(x^2 - x - 6) &= 0 && \text{(common factor } 2\text{)}\\ 2(x - 3)(x + 2) &= 0 && \text{(Sum and Product)}\\ x - 3 = 0 \ &\text{ or } \ x + 2 = 0 && \text{(Null Factor Law: } 2 \neq 0\text{)}\\ x = 3 \ &\text{ or } \ x = -2. \end{aligned}$$ Check: $2 \times 3^2 - 2 \times 3 - 12 = 18 - 6 - 12 = 0$ and $2 \times (-2)^2 - 2 \times (-2) - 12 = 8 + 4 - 12 = 0$. The solutions are $-2$ and $3$.

The Null Factor Law needs one side equal to zero and the other side written as a product: a product equal to a non-zero number, such as $x(x - 3) = 10$, says nothing about its factors. That is why part 2 first brings $3x + 10$ to the left side.

Skills to practice

Solving Equations by Factorising a Trinomial
Writing the Solution of an Equation by Factorising a Trinomial

III Completing the Square

III.1 Solving by Completing the Square

When a quadratic expression does not factorise at sight (a special form, or sum and product), we need a general method. This method is called completing the square. The idea is to rewrite the quadratic as a perfect square plus (or minus) a constant. When solving an equation, this eventually leads to a difference of two squares or to taking square roots.

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Proposition — Completing the Square

Any quadratic expression of the form $x^2 + bx + c$ (leading coefficient $1$) can be put into the form $(x + p)^2 + q$ by completing the square: $$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c.$$ For a general quadratic $ax^2 + bx + c$ with $a\neq 0$, we first take out the factor $a$: $$ax^2 + bx + c = a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right)$$ and then complete the square inside the brackets.

Proof

We start with the expression $x^2 + bx + c$. Our goal is to manipulate the first two terms to create a perfect square trinomial of the form $(x+k)^2 = x^2 + 2kx + k^2$. $$ \begin{aligned} x^2 + bx + c = & (x^2 + bx) + c && \color{gray}\text{(Group the $x$-terms)} \\ \end{aligned} $$ Comparing $x^2+bx$ to $x^2+2kx$, we see that we need $b=2k$, so $k=\frac{b}{2}$. The term needed to complete the square is $k^2 = \left(\frac{b}{2}\right)^2$. We add and subtract this term inside the brackets. $$ \begin{aligned} x^2 + bx + c = & \left(x^2 + bx + \left(\frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2\right) + c && \color{gray}\text{(Add and subtract the missing term)} \\ = & \left(x^2 + bx + \left(\frac{b}{2}\right)^2\right) - \left(\frac{b}{2}\right)^2 + c && \color{gray}\text{(Group the perfect square trinomial)} \\ = & \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c && \color{gray}\text{(Factorise the trinomial)} \end{aligned} $$ This completes the proof.

Example

Complete the square for $x^2 + 10x + 24$.

Answer

We know that $(x + 5)^2 = x^2 + 10x + 25$. So $$\begin{aligned} x^2 + 10x + 24 &= x^2 + 10x + 25 -25 +24 \\ &= (x + 5)^2 - 1 &&\color{gray}\text{(Complete the square)}. \end{aligned}$$

Method — Solving by Completing the Square

Given an equation $ax^2+bx+c=0$ with $a\neq 0$:

Step 1: Complete the square to rewrite the left-hand side as $(x + p)^2$ plus or minus a constant, then the equation in the form $(x + p)^2 = q$ (when $a \neq 1$, first divide both sides by $a$).
Step 2: Conclude from the sign of $q$. If $q > 0$: $x + p = \sqrt{q}$ or $x + p = -\sqrt{q}$ (or, in the same way, factorise the difference of squares $(x + p)^2 - \left(\sqrt{q}\right)^2 = 0$ and apply the Null Factor Law), so there are two solutions. If $q = 0$: $x + p = 0$, so there is one solution, $x = -p$. If $q < 0$: there is no solution, since a square is never negative.
Step 3: Solve the linear equations obtained, and check the solutions in the original equation.

Example

Solve each equation by completing the square. Give the exact solutions.

$x^2 + 6x + 2 = 0$
$x^2 - 4x + 7 = 0$

Answer

Since $(x + 3)^2 = x^2 + 6x + 9$, the first two terms give $x^2 + 6x = (x + 3)^2 - 9$: $$\begin{aligned} x^2 + 6x + 2 &= 0\\ (x + 3)^2 - 9 + 2 &= 0 &&\color{gray}\text{(Complete the square)}\\ (x + 3)^2 &= 7 &&\color{gray}\text{(Add $7$ to both sides)}\\ x + 3 = \sqrt{7} &\text{ or } x + 3 = -\sqrt{7} &&\color{gray}\text{($q = 7 > 0$: two solutions)}\\ x = -3 + \sqrt{7} &\text{ or } x = -3 - \sqrt{7} &&\color{gray}\text{(Solve)}. \end{aligned}$$ Since $7$ is not a perfect square, the trinomial does not factorise at sight, and the solutions stay written with $\sqrt{7}$. Check: $(-3 + \sqrt{7})^2 + 6(-3 + \sqrt{7}) + 2 = 9 - 6\sqrt{7} + 7 - 18 + 6\sqrt{7} + 2 = 0$, and the same computation with $-\sqrt{7}$ also gives $0$. The solutions are $-3 - \sqrt{7}$ and $-3 + \sqrt{7}$.
Since $(x - 2)^2 = x^2 - 4x + 4$, the first two terms give $x^2 - 4x = (x - 2)^2 - 4$: $$\begin{aligned} x^2 - 4x + 7 &= 0\\ (x - 2)^2 - 4 + 7 &= 0 &&\color{gray}\text{(Complete the square)}\\ (x - 2)^2 + 3 &= 0 &&\color{gray}\text{(Simplify)}\\ (x - 2)^2 &= -3 &&\color{gray}\text{(Subtract $3$ from both sides)}. \end{aligned}$$ Here $q = -3 < 0$. A square is never negative, so no value of $x$ makes $(x - 2)^2$ equal to $-3$: the equation has no solution, $S = \varnothing$. Indeed, $x^2 - 4x + 7 = (x - 2)^2 + 3 \geq 3$ for every $x$.

Completing the square works even when the trinomial does not factorise: it gives the exact solutions, or shows that there are none.

Example

Solve $x^2 + 10x + 24 = 0$.

Answer

We know that $(x + 5)^2 = x^2 + 10x + 25$. So $$\begin{aligned} x^2 + 10x + 24 &= 0 \\ x^2 + 10x + 25 -25 +24 &= 0 \\ (x + 5)^2 - 1 &= 0 &&\color{gray}\text{(Complete the square)} \\ (x + 5)^2 - 1^2 &= 0 &&\color{gray}\text{(Difference of squares)}\\ (x + 5 - 1)(x + 5 + 1) &= 0 &&\color{gray}\text{(Factorise)}\\ (x + 4)(x + 6) &= 0 &&\color{gray}\text{(Simplify)}\\ x + 4 = 0 &\text{ or } x + 6 = 0 &&\color{gray}\text{(Null factor law)} \\ x = -4 &\text{ or } x = -6 &&\color{gray}\text{(Solve)}. \end{aligned}$$

Skills to practice

Finding the Solutions by Completing the Square
Solving by Completing the Square

IV Going further

Skills to practice

Reasoning and Explaining
Solving Problems with Quadratic Equations
Investigating Patterns in Quadratic Equations
Quiz