Solve contextual problems involving quadratic equations

Alberta Grade 11 Mathematics

Relations and Functions

Build a model, solve it, and decide which answers fit the situation

A quadratic equation can describe situations involving area, height, or another quantity that depends on a squared value. Solving gives possible values for an unknown. The context helps you decide which values are reasonable answers. In this lesson, you will define an unknown, build and solve a quadratic equation, and check the result.

What you will learn

Represent a contextual situation with a quadratic equation.
Choose a suitable method to solve the equation.
Check proposed roots in the original situation and state answers with units.

1. Review: variables, equations, and units

A variable is a letter that stands for a number. An equation says that two expressions have the same value. A solution makes the equation true.

A quadratic equation has a squared variable as its highest power. A common form is

ax^2+bx+c=0

, where

a

,

b

, and

c

are numbers and

a\ne 0

. For example,

x^2+5x+6=0

is quadratic. The equation

5x+6=0

is not quadratic because its variable is not squared.

Before writing a model, say what the variable measures. Include its unit. If

x

is a length in metres, then

x^2

measures square metres. Units can help reveal a model that does not match the situation.

ax^2+bx+c=0,\quad a\ne 0

A solution makes an equation true.
A quadratic equation has a nonzero coefficient on its squared term.
Define the variable and its unit before building an equation.

2. Build a quadratic model

A model is an equation or relationship that represents a situation. Start by identifying the unknown and the information that connects it to the other quantities.

For a rectangle, area equals length multiplied by width. If its width is

w

metres and its length is

w+3

metres, its area is

w(w+3)

square metres. If the area is known, set this product equal to the given area. Expanding the product gives a quadratic equation.

The quadratic formula solves an equation written in the form

ax^2+bx+c=0

. Substitute the coefficients carefully, including their signs. Factoring is another method when the expression can be rewritten as a product. A square-root method can help when the equation already has a form such as

x^2=k

.

When two factors multiply to zero, at least one factor must be zero. Thus,

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

leads to

x-4=0

or

x+2=0

. The resulting values are called roots: they make the equation true.

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

Translate the situation into an equation before choosing a solving method.
Factoring rewrites an expression as a product.
Use a method that fits the equation you have built.

3. Check roots against the context

A root can be correct algebraically but unsuitable for the situation. A negative length does not describe a side of a garden. A negative time may refer to a moment before an event began, rather than after it.

Substitute each proposed root into the original equation or relationship. Then check the conditions in the question. For example, rectangle side lengths must be positive, and a length left over from a fixed amount of material must also be positive.

Keep units in the final answer. Time uses units such as seconds; length uses units such as metres; area uses square units. Round only when needed, and check that the rounded result still makes sense.

Check each root in the original relationship.
Reject roots that break a condition in the situation.
State the answer in words and include appropriate units.

4. A reliable solution plan

Use this order: define the unknown and its unit, build an equation from the given relationships, solve it, and check the proposed values in the original situation. Then state which value or values answer the question.

This plan helps prevent two common errors: writing an equation that does not match the situation and reporting a root without checking whether it is possible. A complete answer explains what the number represents.

Define, model, solve, check, and state.
A contextual answer must fit both the equation and the situation.

Worked example

Rectangle garden

A rectangular garden is 3 m longer than it is wide. Its area is 40 m². Find its dimensions.

Quadratic model for the garden

The model is $y=w^2+3w-40$. Its zeros are $w=-8$ and $w=5$; only the positive width fits the garden.

Define the unknown
Let $w$ be the width in metres. Then the length is $w+3$ metres.
$\text{length}=w+3$
Build the area equation
Area is width multiplied by length. Set the product equal to the given area, then expand to put the equation in quadratic form.
$w(w+3)=40\;\Longrightarrow\;w^2+3w-40=0$
Factor and solve
The numbers $8$ and $-5$ multiply to $-40$ and add to $3$ . This gives the factorization. Set each factor equal to zero to find the possible widths. (w+8)(w-5)=0 \Longrightarrow w=-8 or w=5
Choose and check
A width cannot be negative, so reject $-8$ . A width of $5$ m gives a length of $8$ m. Their product is the stated area.
$5\times 8=40$

Answer: The garden is 5 m wide and 8 m long.

Check: Both dimensions are positive, differ by 3 m, and have an area of 40 m².

Worked example

A ball reaching the ground

A ball is thrown upward from a platform. Its height above the ground, in metres, after

t

seconds is

h=-5t^2+20t+25

. When does it reach the ground?

Height of the ball

The model is $h=-5t^2+20t+25$. The graph reaches height zero at $t=-1$ and $t=5$; only the positive time fits the question.

Set the height to zero
The ball is on the ground when its height above the ground is zero. Substitute zero for $h$ in the given model.
$-5t^2+20t+25=0$
Simplify and factor
Divide both sides by $-5$ . The equation becomes easier to factor, and dividing both sides by the same nonzero number preserves the solutions.
$t^2-4t-5=0=(t-5)(t+1)$
Find and check the roots
The factors give $t=5$ or $t=-1$ . Negative time is before the ball is thrown, so it does not answer when the ball reaches the ground afterward. At $t=5$ , the original model gives height zero. t=5 or t=-1

Answer: The ball reaches the ground 5 seconds after it is thrown.

Check: Substitution gives

-5(5^2)+20(5)+25=0

. The accepted time is positive.

Worked example

Edging beside a wall

A 10 m strip of edging is used for three sides of a rectangular display beside a straight wall. No edging is needed along the wall. The display area must be 12 m². Find the possible lengths of the two sides perpendicular to the wall.

Area condition for the display

The equation $2x^2-10x+12=0$ has roots $x=2$ and $x=3$. Both give positive side lengths in the stated situation.

Represent the sides
Let $x$ be the length in metres of each side perpendicular to the wall. The side parallel to the wall uses the remaining edging, so its length is $10-2x$ metres.
$2x+(10-2x)=10$
Use the area
The display area is the product of a perpendicular side and the side parallel to the wall. Set that product equal to 12 square metres.
$x(10-2x)=12$
Solve the equation
Rearrange and factor the equation. Each factor gives a possible value for $x$ .
$2x^2-10x+12=0\;\Longrightarrow\;2(x-2)(x-3)=0$
Check both possibilities
For $x=2$ , the side along the wall is $10-2(2)=6$ m. For $x=3$ , it is $10-2(3)=4$ m. Both arrangements have positive sides and the required area.
$2\times6=12,\quad 3\times4=12$

Answer: Each perpendicular side can be 2 m, with a 6 m side along the wall, or 3 m, with a 4 m side along the wall.

Check: Both arrangements use 10 m of edging and have an area of 12 m².

Common mistakes and how to avoid them

Keeping every root as a contextual answer.

Correction: Check each root against the situation. Reject a negative length or a negative time when it does not fit the question.

Using an area value as though it were a side length.

Correction: Area is measured in square units, while a side length is measured in units. Use the area relationship to build the equation.

Not checking a root in the original relationship.

Correction: Substitute the proposed value into the original equation or model and confirm that it gives the stated result.

Lesson summary

Define the variable and its units before writing an equation.
Translate the situation into a quadratic equation and solve it with a suitable method.
Check every root in the original relationship and reject roots that do not fit the situation.
Answer in words and include units and relevant restrictions.

Check your understanding

Question 1

A square garden has area 49 m². Which equation represents its side length

x

?

$x^2=49$
$2x=49$
$x+49=0$
$x^2+49=0$

Show answer and explanation

x^2=49

A square's area is its side length multiplied by itself. The positive side length is 7 m.

Question 2

The equation

(t-6)(t+2)=0

gives possible times in seconds. Which root is a possible time after an event begins?

$t=-2$
$t=2$
$t=6$
$t=8$

Show answer and explanation

t=6

The roots are

6

and

-2

. A negative time is before the event begins, so 6 seconds is the suitable answer.

Question 3

A quadratic model gives roots

x=4

and

x=7

. What should you do before reporting either as an answer?

Check whether each root fits the original context.
Choose the larger root without checking.
Add the roots and report the sum.
Change both roots to negative values.

Show answer and explanation

Check whether each root fits the original context.

A root must make the equation true and make sense in the situation. Check each proposed value before deciding which values to report.

Key terms

Quadratic equation: An equation whose highest power of its variable is 2, with a nonzero coefficient on the squared term.
Root: A value of the variable that makes an equation true.
Factor: An expression multiplied by another expression to make a product.
Model: An equation or relationship used to represent a situation.

Continue through Math 20-2

View the complete Math 20-2 Alberta Grade 11 Mathematics curriculum and lessons

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Working on RF2 · Solve contextual problems involving quadratic equations in Math 20-2? Bring the question and your own attempt. In a live session, you can compare the equation, table and graph, then check what the intercepts or vertex mean. Ask about the first step that does not make sense, then try a similar question yourself.

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برای مبحث RF2 · مسئله‌های کاربردیِ معادلهٔ درجهٔ دوم را حل کنید در درس Math 20-2، سؤال و تلاش خود را همراه بیاورید. در جلسه می‌توانید معادله، جدول و نمودار را مقایسه کنید و معنای ریشه‌ها یا رأس را بررسی کنید. راهنمای معلم ریاضی را ببینید یا جلسهٔ معارفهٔ رایگان ۳۰ دقیقه‌ای درخواست کنید. زمان جلسه پس از پذیرش معلم قطعی می‌شود.

About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 11 Mathematics (Math 20-2), expectation RF2. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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