Quadratic Equations

What is a Quadratic Equation?

A quadratic equation is a mathematical equation involving a quadratic form. This equation contains a variable with the highest power of 2. The general form of a quadratic equation is:

ax^2 + bx + c = 0

with the condition that $a \neq 0$ and $a, b, c$ are real numbers.

Origins of the Term "Quadratic"

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

Quadratic equations can be solved in various ways. Here are some commonly used methods:

Factorization

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

2x^2 - 3x - 2 = 0

(2x + 1)(x - 2) = 0

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

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

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

Completing the Square

This method involves transforming the quadratic equation into a perfect square form.

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}

\left(x - \frac{3}{4}\right)^2 = \frac{16 + 9}{16} = \frac{25}{16}

\left(x - \frac{3}{4}\right)^2 = \left(\frac{5}{4}\right)^2

Therefore:

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

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

x = 2 \text{ or } x = -\frac{1}{2}

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)}

x = \frac{3 \pm \sqrt{9 + 16}}{4}

x = \frac{3 \pm \sqrt{25}}{4}

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

Therefore:

x = \frac{3 + 5}{4} = 2 \text{ or } x = \frac{3 - 5}{4} = -\frac{1}{2}

Formulating Problems as Quadratic Equations

Many real-life problems can be modeled using quadratic equations. Let's explore some examples:

Reading Room Problem

Four reading corners of the same size are created in a classroom measuring 4 m × 6 m. If each corner is a square with side length $x$ meters, then the remaining area of the room for arranging student seating is:

\text{Total area} - \text{Area of four corners}

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

6 m

4 m

x

x

x

x

x

x

x

x

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.

Let's 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

16p - p^2 = 63

-p^2 + 16p - 63 = 0

p^2 - 16p + 63 = 0

By factoring this equation or using the quadratic formula, we can find the values of $p$ and $q$ .

Vehicle Speed Problem

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

Let's say the initial speed is $v$ km/h and the initial travel time is $t$ hours, then:

$v \times t = 320$ (distance = speed × 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

The solution process will result in a quadratic equation that can be solved to find the value of $v$ .

Common Misconceptions About Quadratic Equations

Some common misconceptions include:

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

Concrete example: If a room's length is $x$ meters, and increases by 3 meters, then its length becomes $x + 3$ meters, not $3x$ meters.
Labeling an equation as a quadratic equation simply because the highest power of the variable $x$ is 2, without considering the overall form of the equation.

Remember that a quadratic equation is a polynomial with the standard form $ax^2 + bx + c = 0$ where $a \neq 0$ .

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 $\frac{1}{x}$ .
$\frac{1}{x-5} + \frac{1}{x-3} = x^2 - 4$
This is not a quadratic equation in standard form, because it has a fractional form with variables in the denominator.
$3x^2 + 2x - 1 = 0$
This is a quadratic equation because it is in the form $ax^2 + bx + c = 0$ with $a = 3 \neq 0$ .
$x^2 + \frac{1}{x} + 3 = 0$
This is not a quadratic equation because it contains the term $\frac{1}{x}$ .

Practice Problems

Identifying Quadratic Equations

Determine whether the following mathematical equations are quadratic equations:

$x^3 + x^2 + x = 0$
$3x^2 + 2x - 1 = 0$
$\frac{1}{x} + 5x = 0$
$x^2 + \frac{1}{x} + 3 = 0$

Factorization

Expand the following equations:

$(x + 2)(x + 3) = 0$
$\left(\frac{1}{3}x - 4\right)(x + 9) = 0$
$(2x - 8)(x + 5) = 0$

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.
$3x^2 + 2x - 1 = 0$
Answer: Quadratic equation, because it is in the form $ax^2 + bx + c = 0$ with $a = 3$ , $b = 2$ , and $c = -1$ .
$\frac{1}{x} + 5x = 0$
Answer: Not a quadratic equation, because it contains the term $\frac{1}{x}$ . This is a fractional equation.
$x^2 + \frac{1}{x} + 3 = 0$
Answer: Not a quadratic equation, because it contains the term $\frac{1}{x}$ . This is a mixed equation.

Factorization

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

$(x + 2)(x + 3) = 0$
$x^2 + 3x + 2x + 6 = 0$
$x^2 + 5x + 6 = 0$

So, the product of the two factors is $x^2 + 5x + 6 = 0$
$\left(\frac{1}{3}x - 4\right)(x + 9) = 0$
Answer:

$\left(\frac{1}{3}x - 4\right)(x + 9) = 0$
$\frac{1}{3}x \cdot x + \frac{1}{3}x \cdot 9 - 4 \cdot x - 4 \cdot 9 = 0$
$\frac{1}{3}x^2 + 3x - 4x - 36 = 0$
$\frac{1}{3}x^2 - x - 36 = 0$

Multiply all terms by 3 to simplify:

$x^2 - 3x - 108 = 0$

So, the product of the two factors is $x^2 - 3x - 108 = 0$ or $\frac{1}{3}x^2 - x - 36 = 0$
$(2x - 8)(x + 5) = 0$
Answer:

$(2x - 8)(x + 5) = 0$
$2x \cdot x + 2x \cdot 5 - 8 \cdot x - 8 \cdot 5 = 0$
$2x^2 + 10x - 8x - 40 = 0$
$2x^2 + 2x - 40 = 0$

Factoring $2x - 8$ as $2(x - 4)$ :

$2(x - 4)(x + 5) = 0$

So, the product of the two factors is $2x^2 + 2x - 40 = 0$

Solving Quadratic Equations

Let's solve some equations from the factorization results above:

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

$x^2 + 5x + 6 = 0$
$(x + 2)(x + 3) = 0$

Therefore:
- If $x + 2 = 0$ , then $x = -2$
- If $x + 3 = 0$ , then $x = -3$
The roots of the equation are: $x = -2$ or $x = -3$
$x^2 - 3x - 108 = 0$
Answer: Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{3 \pm \sqrt{9 + 432}}{2}$
$x = \frac{3 \pm \sqrt{441}}{2}$
$x = \frac{3 \pm 21}{2}$

Therefore:
- $x = \frac{3 + 21}{2} = 12$
- $x = \frac{3 - 21}{2} = -9$
The roots of the equation are: $x = 12$ or $x = -9$

Verification by factorization:

$x^2 - 3x - 108 = 0$
$(x - 12)(x + 9) = 0$
$2x^2 + 2x - 40 = 0$
Answer: Simplify the equation by dividing all terms by 2:

$x^2 + x - 20 = 0$

Factorization:

$x^2 + x - 20 = 0$
$(x + 5)(x - 4) = 0$

Therefore:
- If $x + 5 = 0$ , then $x = -5$
- If $x - 4 = 0$ , then $x = 4$
The roots of the equation are: $x = -5$ or $x = 4$

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