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 $a \neq 0$ .

A square with side length $x$ has area $x\cdot x=x^2$ . Doubling the side makes the area $four times as large$ , not twice as large. For example, a side of 2 meters gives an area of 4 square meters, while a side of 4 meters gives 16 square meters.

How to Solve Quadratic Equations

The form of the equation determines the $most convenient method$ . Each method below works when its calculation conditions are met. 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:

Split the middle term into $-4x+x$ so both groups have $a common factor$ . Once the product equals zero, at least one factor must be zero:

\begin{aligned}2x^2-3x-2&=2x^2-4x+x-2\\&=2x(x-2)+(x-2)\\&=(2x+1)(x-2)\end{aligned}

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

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

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 $quadratic formula$ :

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

6 \text{ m}

4 \text{ m}

If the remaining area must be $20 square meters$ , the area model becomes an equation. The four drawn squares show the arrangement schematically. Calculate their side length from the area requirement:

\begin{aligned}24-4x^2&=20\\4x^2&=4\\x^2&=1\\x&=\pm1\end{aligned}

A side length must be positive, so $x=1$ meter. The two squares along the 4-meter wall must not overlap, requiring $2x\le4$ . One meter satisfies this restriction.

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 $3x$ would multiply the original length by three.
Labeling an equation as quadratic solely because it contains the term $x^2$ , without considering the equation's overall form.

A quadratic equation $is a polynomial equation of the form$ $ax^2 + bx + c = 0$ with $a \neq 0$ .

Forms of Quadratic Equations

Distinguish an equation $as written$ from the form obtained after clearing denominators. Multiplying by a denominator must preserve the restriction that the denominator cannot be zero.

Original equation	After clearing denominators	Classification
$\frac1x+2x+4=0$	$2x^2+4x+1=0,\ x\ne0$	Rational, reducible to a quadratic for permitted x values
$\frac1{x-5}+\frac1{x-3}=x^2-4$	$2x-8=(x^2-4)(x-5)(x-3),\ x\ne3,5$	Rational, leads to a degree-four equation
$3x^2+2x-1=0$	$a=3,\ b=2,\ c=-1$	Already a quadratic equation
$x^2+\frac1x+3=0$	$x^3+3x+1=0,\ x\ne0$	Rational, leads to a cubic equation

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:

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

$Expanding Products$

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

For classification, inspect the $highest power and any denominators$ . To expand, multiply each term in the first factor by each term in the second. To find roots, use a product equal to zero.

$Classification problem 1$ : $x^3 + x^2 + x = 0$

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

$Classification problem 2$ : $3x^2 + 2x - 1 = 0$

$Answer$ : Quadratic equation, because it is in the form $ax^2 + bx + c = 0$ with $a = 3$ , $b = 2$ , and $c = -1$ .

$Classification problem 3$ : $\frac{1}{x} + 5x = 0$

$Answer$ : Not a quadratic equation, because it contains the term $\frac{1}{x}$ . As written, it is rational. For $x\ne0$ , multiplying by x gives $1+5x^2=0$ .

$Classification problem 4$ : $x^2 + \frac{1}{x} + 3 = 0$

$Answer$ : Not a quadratic equation, because it contains the term $\frac{1}{x}$ . This is a rational equation.

$Expansion problem 1$ : $(x + 2)(x + 3) = 0$

$Answer$ :

(x + 2)(x + 3) = 0

So, the product of the two factors is $x^2 + 5x + 6 = 0$

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

$Answer$ :

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

$Expansion problem 3$ : $(2x - 8)(x + 5) = 0$

$Answer$ :

(2x - 8)(x + 5) = 0

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

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

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

Solve some equations from the factorization results above:

$Roots for expansion problem 1$ : $x^2 + 5x + 6 = 0$

$Answer$ : Factorization:

x^2 + 5x + 6 = 0

Set each factor equal to zero:

If $x + 2 = 0$ , then $x = -2$
If $x + 3 = 0$ , then $x = -3$

The roots of the equation are: $x = -2$ or $x = -3$

$Roots for expansion problem 2$ : $x^2 - 3x - 108 = 0$

$Answer$ : Using the quadratic formula:

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

The two signs in the quadratic formula give:

$x = \frac{3 + 21}{2} = 12$
$x = \frac{3 - 21}{2} = -9$

The roots of the equation are: $x = 12$ or $x = -9$

Verification by factorization:

x^2 - 3x - 108 = 0

$Roots for expansion problem 3$ : $2x^2 + 2x - 40 = 0$

$Answer$ : Simplify the equation by dividing all terms by $2$ :

x^2 + x - 20 = 0

Factorization:

x^2 + x - 20 = 0

Set each factor equal to zero:

If $x + 5 = 0$ , then $x = -5$
If $x - 4 = 0$ , then $x = 4$

The roots of the equation are: $x = -5$ or $x = 4$

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 $a \neq 0$ .

Squares and Areas

How to Solve Quadratic Equations

Factorization

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

Split the middle term into $-4x+x$ so both groups have $a common factor$ . Once the product equals zero, at least one factor must be zero:

\begin{aligned}2x^2-3x-2&=2x^2-4x+x-2\\&=2x(x-2)+(x-2)\\&=(2x+1)(x-2)\end{aligned}

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

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

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 $quadratic formula$ :

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

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

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

6 \text{ m}

4 \text{ m}

\begin{aligned}24-4x^2&=20\\4x^2&=4\\x^2&=1\\x&=\pm1\end{aligned}

A side length must be positive, so $x=1$ meter. The two squares along the 4-meter wall must not overlap, requiring $2x\le4$ . One meter satisfies this restriction.

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

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

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 $3x$ would multiply the original length by three.
Labeling an equation as quadratic solely because it contains the term $x^2$ , without considering the equation's overall form.

A quadratic equation $is a polynomial equation of the form$ $ax^2 + bx + c = 0$ with $a \neq 0$ .

Forms of Quadratic Equations

Distinguish an equation $as written$ from the form obtained after clearing denominators. Multiplying by a denominator must preserve the restriction that the denominator cannot be zero.

Original equation	After clearing denominators	Classification
$\frac1x+2x+4=0$	$2x^2+4x+1=0,\ x\ne0$	Rational, reducible to a quadratic for permitted x values
$\frac1{x-5}+\frac1{x-3}=x^2-4$	$2x-8=(x^2-4)(x-5)(x-3),\ x\ne3,5$	Rational, leads to a degree-four equation
$3x^2+2x-1=0$	$a=3,\ b=2,\ c=-1$	Already a quadratic equation
$x^2+\frac1x+3=0$	$x^3+3x+1=0,\ x\ne0$	Rational, leads to a cubic equation

Practice Problems

$Identifying Quadratic Equations$

Determine whether the following mathematical equations are quadratic equations:

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

$Expanding Products$

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

For classification, inspect the $highest power and any denominators$ . To expand, multiply each term in the first factor by each term in the second. To find roots, use a product equal to zero.

$Classification problem 1$ : $x^3 + x^2 + x = 0$

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

$Classification problem 2$ : $3x^2 + 2x - 1 = 0$

$Answer$ : Quadratic equation, because it is in the form $ax^2 + bx + c = 0$ with $a = 3$ , $b = 2$ , and $c = -1$ .

$Classification problem 3$ : $\frac{1}{x} + 5x = 0$

$Answer$ : Not a quadratic equation, because it contains the term $\frac{1}{x}$ . As written, it is rational. For $x\ne0$ , multiplying by x gives $1+5x^2=0$ .

$Classification problem 4$ : $x^2 + \frac{1}{x} + 3 = 0$

$Answer$ : Not a quadratic equation, because it contains the term $\frac{1}{x}$ . This is a rational equation.

$Expansion problem 1$ : $(x + 2)(x + 3) = 0$

$Answer$ :

(x + 2)(x + 3) = 0

So, the product of the two factors is $x^2 + 5x + 6 = 0$

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

$Answer$ :

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

$Expansion problem 3$ : $(2x - 8)(x + 5) = 0$

$Answer$ :

(2x - 8)(x + 5) = 0

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

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

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

Solve some equations from the factorization results above:

$Roots for expansion problem 1$ : $x^2 + 5x + 6 = 0$

$Answer$ : Factorization:

x^2 + 5x + 6 = 0

Set each factor equal to zero:

If $x + 2 = 0$ , then $x = -2$
If $x + 3 = 0$ , then $x = -3$

The roots of the equation are: $x = -2$ or $x = -3$

$Roots for expansion problem 2$ : $x^2 - 3x - 108 = 0$

$Answer$ : Using the quadratic formula:

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

The two signs in the quadratic formula give:

$x = \frac{3 + 21}{2} = 12$
$x = \frac{3 - 21}{2} = -9$

The roots of the equation are: $x = 12$ or $x = -9$

Verification by factorization:

x^2 - 3x - 108 = 0

$Roots for expansion problem 3$ : $2x^2 + 2x - 40 = 0$

$Answer$ : Simplify the equation by dividing all terms by $2$ :

x^2 + x - 20 = 0

Factorization:

x^2 + x - 20 = 0

Set each factor equal to zero:

If $x + 5 = 0$ , then $x = -5$
If $x - 4 = 0$ , then $x = 4$

The roots of the equation are: $x = -5$ or $x = 4$