# Equations

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 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. $$$ C = \frac{5(F-32)}{9}$$$

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 $$-1$$. |

$$x-1=0$$ | $$1$$ | $$1$$ | $$\Tblue{1}-1=0$$ |

$$x^2=1$$ | $$2$$ | $$-1$$, $$1$$ | $$\Tblue{1}^2 = (\Tblue{-1})^2=1$$ |

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 $$\Tgreen{c}$$.

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 $$$\displaystyle \Tred{t} = \frac{-4-8}{5} = -\frac{12}{5}.$$$

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 $$\Tblue{c}$$.

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*}

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 | $$\Tviolet{\Delta}\lt 0$$ | $$\Tviolet{\Delta}= 0$$ | $$\Tviolet{\Delta}\gt 0$$ |
---|---|---|---|

Formula of solution | No solution | $$\displaystyle\frac{-\Tblue{b}}{2\Tblue{a}}$$ | $$\displaystyle\frac{-\Tblue{b}\pm\sqrt{\Tviolet{\Delta}}}{2\Tblue{a}}$$ |

Number of solutions | $$0$$ | $$1$$ | $$2$$ |

Example | $$\Tred{x}^2 + 1 = 0$$ | $$\Tred{x}^2 - 2\Tred{x} + 1 = 0$$ | $$\Tred{x}^2 +2\Tred{x} = 0$$ |

$$\Delta = $$ | $$-1$$ | $$0$$ | $$4$$ |

Solutions | None | $$1$$ | $$-2$$, $$0$$ |