Use adaptive quiz-based learning to study this topic faster and more effectively.

# Equations

## Definition of an equation

An equation is an equality that involves one or several variables.

Here are several equations \begin{align*} &\Tred{x}-1 \Tblue{=} 0,\quad (\Tred{t}-1)(\Tred{t}-2)\Tblue{=}4,\quad \frac{\Tred{a}+1}{\Tred{a}^2-\Tred{a}} \Tblue{=} 2\\ &\qquad \Tred{x} + \Torange{y} \Tblue{=} 1,\quad \Tred{x}^2 + \Torange{y}^2 \Tblue{=} 1. \end{align*}

The variables in the equation are called the unknowns.

$\Tred{a}$ is an unknown in the equation $(\Tred{a}+1)(\Tred{a}-1) = 0$.

An equation is like a scale where two expressions need to be equal.

## Rearranging equations

An equation or formula can be rearranged to change its subject.

The formula for the temperature in Fahrenheit, $\Tred{F}$, in terms of the temperature in Celsius, $\Tblue{C}$ is $$\Tred{F} = \dfrac{9\Tblue{C}}{5} +32.$$ If we want to find $\Tblue{C}$ in terms of $\Tred{F}$, we must rearrange the equation.

To rearrange an equation, we must always do the same to both sides of an equation.

When we start to rearrange $F = 9C/5 +32$, we first subtract $\Tblue{32}$ from both sides. Now we have $F-\Tblue{32} = 9C/5$.

Now we multiply both sides by $\Tgreen{5}$ to get $\Tgreen{5} (F-32)=9C$. Finally we divide both sides by $\Tviolet{9}$ to give us the fully rearranged formula. $$C = \frac{5(F-32)}{9}$$

## Solution of an equation

A solution of an equation is a value for the unknown that satisfies the equation.

$\Tblue{-1}$ the a solution of the equation $\Tred{x} +1 =0$ because $$(\Tblue{-1}) + 1 = 0.$$ $\Tgreen{0}$ is not a solution because $$(\Tgreen{0}) + 1 = 1 \ne 0.$$

The solution is also called the root or the zero of the equation.

An equation can have one, several or no solution.

Equation Number of solutions Solutions Reason
$x^2=-1$ $0$ None $x^2$ is always positive and so no number can be squared to equal $-1$.
$x-1=0$ $1$ $1$ $\Tblue{1}-1=0$
$x^2=1$ $2$ $-1$, $1$ $\Tblue{1}^2 = (\Tblue{-1})^2=1$

## Linear equations

A linear equation is an equation that depends only on the variable and not on any power. It only involves linear functions of the variable.

The following equations are linear: $$\Tred{x}-1 = 0, \quad 3\Tred{ t } + 2 = -2, \quad \frac{\Tred{y}}{2} + \frac{4}{5} = -\frac{\sqrt{3}}{7}$$

The following equations are not linear: $$\Tred{x}^2 = 1,\quad \sqrt{\Tred{t}} - 1 = 0,\quad \frac{1}{\Tred{y}} + \frac{1}{\Tred{y}-1} = 1$$

The general form of a linear equation is $$\Tgreen{a} \Tred{x} + \Tgreen{b} = \Tgreen{c}$$ for numbers $\Tgreen{a}\ne0$, $\Tgreen{b}$ and $\Tgreen{c}$.

## Solutions of linear equations

We want to solve a general linear equation $$\Tblue{a} \Tred{x} + \Tblue{b} = \Tblue{c}$$ for numbers $\Tblue{a}\ne0$, $\Tblue{b}$ and $\Tblue{c}$. The unknown is $\Tred{x}$.

The equation has a unique solution and it is the fraction $$\Tred{x} = \frac{\Tblue{c-b}}{\Tblue{a}}.$$

The solution of the equation $2\Tred{x}-1=3$ is $$\displaystyle \Tred{x} = \frac{3-(-1)}{2} = \frac{4}{2} = 2.$$ This can be checked by putting the solution back into the equation $$2\times \Tred{2}- 1 = 4-1 = 3.$$

The solution of the equation $5\Tred{t}+ 8 = - 4$ is $$\displaystyle \Tred{t} = \frac{-4-8}{5} = -\frac{12}{5}.$$

## Solving a linear equation

We solve a linear equation by applying the same operations to both sides of the expression. The objective is to isolate the unknown $x$.

Let's start with an example. We want to solve $$2\Tred{x}-1 = 3.$$

• We add $\Tgreen{1}$ to both sides $$2\Tred{x} = 2\Tred{x}-1 + \Tgreen{1} = 3+\Tgreen{1} = 4$$
• We divide both sides by $\Tgreen{2}$ $$\Tred{x} = \frac{2\Tred{x}}{\Tgreen{2}} = \frac{4}{\Tgreen{2}} = 2$$
• This means that the solution is $\Tred{x} = 2$.

We can replicate the reasoning to solve a general linear equation $$\Tblue{a} \Tred{x} + \Tblue{b} = \Tblue{c}$$ for numbers $\Tblue{a}\ne0$, $\Tblue{b}$ and $\Tblue{c}$. This gives the solution $\displaystyle \Tred{x} = \frac{\Tblue{c-b}}{\Tblue{a}}.$

• We subtract $\Tgreen{b}$ from both sides: $\Tblue{a}\Tred{x} = \Tblue{a}\Tred{x} + \Tblue{b} - \Tgreen{b} = \Tblue{c} - \Tgreen{b}$
• We divide both sides by $\Tgreen{a}$: $\displaystyle \Tred{x} = \frac{\Tblue{a}\Tred{x}}{\Tgreen{a}} = \frac{\Tblue{c-b}}{\Tgreen{a}}.$

A quadratic equation is an equation that depends only on the variable and its square. It only involves quadratic functions of the variable.

Quadratic equations may or may not have a linear term and a constant. They are often written so that the right hand side is zero.

The following equations are quadratic: $$\Tred{x^2}-1 = 0, \quad 3\Tred{t^2} + 2\Tred{t} + 1 = -2.$$

The following equations are not quadratic: $$\Tred{x}^3 = 1,\quad \sqrt{\Tred{t}} - 1 = 0,\quad \frac{1}{\Tred{y}+1} + \frac{1}{\Tred{y}} + \frac{1}{\Tred{y}-1} = 1$$

The general form of a quadratic equation is $$\Tblue{a} \Tred{x}^2 + \Tblue{b}\Tred{x} + \Tblue{c} = 0$$ for numbers $\Tblue{a}\ne0$, $\Tblue{b}$ and $\Tblue{c}$.

## Discriminant of a quadratic equation

The discriminant $\Tviolet{\Delta}$ of a quadratic equation $$a\Tred{x}^2 + b\Tred{x}+c = 0$$ is the number $$\Tviolet{\Delta} = b^2 - 4 ac.$$ The computation of the discriminant is essential for solving a quadratic equation.

The discriminant of the equation $\Tred{x}^2 - 3\Tred{x} + 2 = 0$ is $$\Tviolet{\Delta} = 3^2 - 4\times 1\times 2 = 9-8 = 1.$$

The discriminant can be positive, negative or zero. The sign of the discriminant tells us whether an equation can be solved.

Equation Discriminant Sign Number of solutions
$x^2 - 3x + 1 = 0$ $9 - 4 = 5$ $\gt 0$ $2$
$x^2 - 2x + 1 = 0$ $4 - 4 = 0$ $= 0$ $1$
$x^2 - x + 1 = 0$ $1 - 4 = -3$ $\lt 0$ $0$

We want to solve a quadratic equation $$a\Tred{x}^2 + b\Tred{x}+c = 0$$ for numbers $a\ne0$, $b$ and $c$.

We first compute the discriminant $\Tviolet{\Delta} = b^2 - 4 ac$.

• If $\Tviolet{\Delta}\lt 0$, the equation has no solutions.
• If $\Tviolet{\Delta} = 0$, the equation has exactly one solution $$\Tred{x_0} = \frac{-b}{2a}.$$
• If $\Tviolet{\Delta}\gt 0$, the equation has exactly two solutions $$\displaystyle \Tred{x_-} = \frac{-b-\sqrt{\Tviolet{\Delta}}}{2a},\quad \Tred{x_+} = \frac{-b+\sqrt{\Tviolet{\Delta}}}{2a}.$$

For instance, take the equation $$x^2 - x - 2 = 0$$ The discriminant is $\Tviolet{\Delta} = 1 + 8 = 9$. Its positive square root is $\sqrt{\Tviolet{\Delta}} = \Tgreen{3}$. The solutions of the equation are \begin{align*} \Tred{x_-} &= \frac{-(-1) - \Tgreen{3}}{2\times 1} = \frac{-2}{2} = -1\\ \Tred{x_+} &= \frac{-(-1) + \Tgreen{3}}{2\times 1} = \frac{4}{2} = 2 \end{align*} This can be checked by plugging the solutions into the equation \begin{align*} (\Tred{-1})^2 - (\Tred{-1}) -2 &= 1 +1 -2 = 0,\\ \Tred{2}^2 -\Tred{2} - 2 &= 4 - 2 - 2 = 0. \end{align*}

## Solving a quadratic equation (overview)

Summary of the solutions for the quadratic equation $$\Tblue{a}\Tred{x}^2 + \Tblue{b}\Tred{x}+\Tblue{c} = 0$$ for numbers $\Tblue{a}\ne0$, $\Tblue{b}$ and $\Tblue{c}$.

We use the discriminant $\Tviolet{\Delta} = \Tblue{b}^2 - 4 \Tblue{ac}$.

Case Formula of solution Number of solutions $\Tviolet{\Delta}\lt 0$ $\Tviolet{\Delta}= 0$ $\Tviolet{\Delta}\gt 0$ No solution $\displaystyle\frac{-\Tblue{b}}{2\Tblue{a}}$ $\displaystyle\frac{-\Tblue{b}\pm\sqrt{\Tviolet{\Delta}}}{2\Tblue{a}}$ $0$ $1$ $2$ $\Tred{x}^2 + 1 = 0$ $\Tred{x}^2 - 2\Tred{x} + 1 = 0$ $\Tred{x}^2 +2\Tred{x} = 0$ $-1$ $0$ $4$ None $1$ $-2$, $0$