Quadratic Equations

Equations of Degree Two

A quadratic equation is a polynomial equation in one variable: it contains $x^2$ but no higher power of $x$ . The general form of a quadratic equation is:

ax^2 + bx + c = 0

In this form, $a, b, c$ are real numbers and $a \neq 0$ .

The term "quadratic" comes from the Latin word quadratus, which means "to make a square." This relates to the geometric interpretation of the form $x^2$ which can be viewed as the area of a square with side length $x$ .

How to Solve Quadratic Equations

No single method fits every quadratic equation, so the choice depends on the form of the equation. Factorisation is quickest when the factors are whole numbers, the formula always works, and completing the square also reveals the vertex.

Factorization

The factorization method involves breaking down the quadratic equation into a product of two linear factors. For example:

2x^2 - 3x - 2 = 0

From the factored form above, we can get the solutions:

If $2x + 1 = 0$ , then $x = -\frac{1}{2}$
If , then

The roots of the quadratic equation are $x = -\frac{1}{2}$ or $x = 2$ .

Completing the Square

Completing the square rewrites the quadratic expression as a squared binomial plus a constant.

Example:

2x^2 - 3x - 2 = 0

We divide all terms by $2$ :

x^2 - \frac{3}{2}x - 1 = 0

Move the constant to the right side:

x^2 - \frac{3}{2}x = 1

Add $\left(\frac{-3/2}{2}\right)^2 = \frac{9}{16}$ to both sides:

x^2 - \frac{3}{2}x + \frac{9}{16} = 1 + \frac{9}{16}

The calculation gives:

x - \frac{3}{4} = \pm \frac{5}{4}

Using the Quadratic Formula

For the equation $ax^2 + bx + c = 0$ , the roots can be determined using the formula:

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

Example:

2x^2 - 3x - 2 = 0

With $a = 2$ , $b = -3$ , and $c = -2$ :

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

The calculation gives:

\begin{aligned} &x = \frac{3 + 5}{4} = 2 \\ &\text{or} \\ &x = \frac{3 - 5}{4} = -\frac{1}{2} \end{aligned}

Formulating Problems as Quadratic Equations

Word problems hide the equation inside a description, so translate each sentence into a quantity first. Only once the equation stands on its own do you choose a solving method.

Quadratic equations model quantities whose relationships contain products or squares. The following examples turn those relationships into equations.

Reading Room Problem

Four reading corners of the same size are created in a room measuring $4 \text{ m} \times 6 \text{ m}$ . If each corner is a square with side length $x \text{ meters}$ , then the remaining area of the room for arranging seats is:

Subtract the area of the four corners from the total area.

4 \times 6 - 4x^2 = 24 - 4x^2

Problem of Multiplying Two Numbers

The product of two numbers is $63$ and their sum is $16$ . We can solve this using a quadratic equation.

Say the two numbers are $p$ and $q$ , then:

$p + q = 16$ , so $q = 16 - p$
$p \times q = 63$

Substituting the value of $q$ :

p(16 - p) = 63

Factor the quadratic equation:

p^2 - 16p + 63 = 0

The two numbers are $7$ and $9$ .

Vehicle Speed Problem

A vehicle travels a distance of $320 \text{ km}$ at a certain speed. If the vehicle travels $24 \text{ km/h}$ faster, its travel time is reduced by $3 \text{ hours}$ . We can find the initial speed using a quadratic equation.

Say the initial speed is $v \text{ km/h}$ and the initial travel time is $t \text{ hours}$ , then:

$v \times t = 320$ (distance $=$ speed $\times$ time)
$(v + 24) \times (t - 3) = 320$ (second condition)

From the first equation: $t = \frac{320}{v}$

Substituting into the second equation:

(v + 24) \times \left(\frac{320}{v} - 3\right) = 320

Multiply by $v$ and simplify:

(v + 24)(320 - 3v) = 320v

A negative speed is not meaningful in this context. Therefore, the initial speed is $v = 40 \text{ km/h}$ . The initial trip takes $8 \text{ hours}$ . At $64 \text{ km/h}$ , the trip takes $5 \text{ hours}$ , exactly $3 \text{ hours}$ less.

Common Misconceptions About Quadratic Equations

Some common misconceptions include:

Identifying the addition operation $x + 3$ as $3x$ .

Concrete example: If a room is $x \text{ meters}$ long and gains $3 \text{ meters}$ , its new length is $(x + 3) \text{ meters}$ . The expression would multiply the original length by three.

Forms of Quadratic Equations

Consider the following forms, which ones are quadratic equations?

$\frac{1}{x} + 2x + 4 = 0$
This is not a quadratic equation because it contains the term $\frac{1}{x}$ .

Practice Problems

These problems mix several shapes, so decide for each equation whether it factors, whether the formula is needed, or whether a perfect square is already there. Write the chosen method down before you calculate.

Identifying Quadratic Equations

Determine whether the following mathematical equations are quadratic equations:

$x^3 + x^2 + x = 0$
$3x^2 + 2x - 1 = 0$

Factorization

Expand the following equations:

$(x + 2)(x + 3) = 0$
$\left(\frac{1}{3}x - 4\right)(x + 9) = 0$

Answer Key

Identifying Quadratic Equations

$x^3 + x^2 + x = 0$
Answer: Not a quadratic equation, because it has the highest power of $3$ ( $x^3$ ). This is a cubic equation.

Factorization

$(x + 2)(x + 3) = 0$
Answer:
$(x + 2)(x + 3) = 0$

Solving Quadratic Equations

Solve some equations from the factorization results above:

$x^2 + 5x + 6 = 0$
Answer: Factorization:
$x^2 + 5x + 6 = 0$