A3.3 Solve real-world problems from quadratic equations

Ontario Grade 11 Mathematics

Quadratic Functions

Turn a situation into an equation, solve it, and decide what the solutions mean.

A quadratic equation can model a situation where a quantity is multiplied by another quantity that depends on it. For example, the area of a rectangle depends on its length multiplied by its width. Solving a real-world problem means more than finding numbers: you must connect the numbers to the story and decide whether each answer is reasonable. In this lesson, you will build an equation from a situation, solve it, and interpret the result.

What you will learn

Identify a quantity that can be represented by a variable.
Translate a real-world relationship into a quadratic equation.
Solve a quadratic equation using factoring when it is suitable.
Check solutions and choose those that make sense in the situation.

1. Prerequisite bridge: quantities and equations

A variable is a letter that stands for an unknown number. In a word problem, choose a variable for the quantity you need to find. State what it represents and include its unit when one is given.

An equation says that two expressions have the same value. To solve an equation, find the value or values of the variable that make the equation true. For example, if a rectangle's width is

x

metres and its length is

x+3

metres, multiplying the dimensions gives its area in square metres.

A quadratic equation is an equation that can be written with a squared variable, such as

x^2+3x-40=0

. In real-world problems, the squared term often comes from multiplying two dimensions that both depend on the unknown.

A=lw

Define your variable before writing the equation.
Keep units attached to the quantities they describe.
A quadratic equation includes a variable squared.

2. From a situation to a quadratic model

A mathematical model is an equation that represents the important relationships in a situation. To create one, identify the unknown, express other quantities using that unknown, and use the information given in the problem.

Consider a rectangular patio whose length is

3

metres greater than its width. If the width is

x

metres, its length is

x+3

metres. If its area is

40

square metres, the area relationship gives

x(x+3)=40

. This is quadratic because multiplying

x

by

x+3

creates a squared term.

To solve, first write the equation in a form with zero on one side. Then choose a method that fits the equation. Factoring is useful when the quadratic expression can be written as a product of two simpler expressions. If a product is zero, at least one factor must be zero. This lets you solve two simpler equations.

A context can restrict which answers are acceptable. A negative number may solve the equation, but it cannot represent a length. Check each candidate against both the original equation and the situation.

x(x+3)=40

Translate each stated relationship into an expression or equation.
For factoring, rewrite the equation so one side is zero.
An algebraic solution is not automatically a sensible real-world answer.

3. Solve, check, and interpret

Work in a clear order: define the variable, build the equation, solve it, and interpret the results. Do not stop as soon as you obtain values for the variable. State the answer using the quantity and units from the problem.

A solution can be checked in two ways. Substitute it into the original equation to confirm that the equation is true. Then compare it with the story: check units, size, and any limits such as a length needing to be positive.

Sometimes a quadratic situation has two solutions that both fit the mathematics. The story determines whether both are meaningful. For example, two possible times may both be relevant in one situation, while a negative length must be rejected in a dimensions problem.

Check candidate values in the original relationship.
Use the context to decide which values are meaningful.
Give a sentence answer with appropriate units.

4. A reliable problem-solving routine

Before calculating, make sure your equation matches the information in the problem. A quick sketch or a short list of known and unknown quantities can help. For a rectangle, label its dimensions and connect area to length multiplied by width.

After solving, compare the answer with the situation. If the answer is a length, it should be expressed in length units. If the question asks for area, use square units. This final interpretation is part of solving the problem, not an optional extra.

Technology can help check arithmetic or display a graph, but you still need to explain what the solutions mean. A graph's horizontal intercepts correspond to values that make the quadratic expression equal to zero. Use the original context to decide which intercepts answer the question.

Represent the quantities before choosing a solving method.
Use units consistently.
Explain why a solution is accepted or rejected.

Turning the patio information into mathematics

Information	Representation
Width	$x$ metres
Length is 3 metres greater	$x+3$ metres
Area is 40 square metres	$x(x+3)=40$
Positive solution and matching length	$x=5$ , so the length is $8$ metres

Worked example

Finding the dimensions of a patio

A rectangular patio is

3

metres longer than it is wide. Its area is

40

square metres. Find its width and length.

Choose a variable
Let $x$ be the patio's width in metres. Since the length is $3$ metres greater, represent it as $x+3$ metres.
$w=x, l=x+3$
Build the equation
The area of a rectangle is width multiplied by length. Substitute the expressions for the dimensions and set the area equal to $40$ square metres.
$x(x+3)=40$
Write the equation in zero form
Expand the product and subtract $40$ from both sides. Keeping zero on one side prepares the equation for factoring.
$x^2+3x-40=0$
Factor and solve
The numbers $8$ and $-5$ multiply to $-40$ and add to $3$ . Use them to factor the quadratic. A product equals zero when at least one factor equals zero, so solve each resulting equation. (x+8)(x-5)=0, x=-8 or x=5
Check the context
The value $x=-8$ would give a negative width, so it cannot describe a patio. For $x=5$ , the length is $5+3=8$ metres, and the area is $5\times8=40$ square metres.
$5\times8=40$

Answer: The patio is

5

metres wide and

8

metres long.

Check: Both dimensions are positive, the length is

3

metres greater than the width, and their product is

40

square metres.

Common mistakes and how to avoid them

Writing the length as

3x

when it is 3 metres greater than the width.

Correction: “3 greater” means add

3

, so the length is

x+3

. A multiple such as

3x

would mean three times the width.

Forgetting to move all terms to one side before factoring.

Correction: Write the equation with zero on one side first. For the patio, use

x^2+3x-40=0

.

Reporting every algebraic solution as a real-world answer.

Correction: Check whether each value makes sense in the situation. A negative width is not a possible patio dimension.

Giving dimensions without units or without checking the area.

Correction: State the dimensions in metres and verify that their product is the stated area in square metres.

Lesson summary

Choose and define a variable for the unknown quantity.
Use the relationships in the problem to build a quadratic equation.
Solve the equation and check candidate values in the original relationship.
Interpret the solutions using the situation and its units.

Check your understanding

Question 1

A rectangle has width

x

metres and length

x+2

metres. Which equation represents an area of

24

square metres?

$x(x+2)=24$
$x+x+2=24$
$2x=24$
$x(x+2)=0$

Show answer and explanation

x(x+2)=24

Area is width multiplied by length. Substituting the dimensions gives

x(x+2)

, which must equal

24

.

Question 2

The equation for a situation factors as

(x-6)(x+4)=0

. The variable represents a length. Which value is meaningful?

$x=-4$
$x=4$
$x=6$
$x=10$

Show answer and explanation

x=6

The factors give

x=6

or

x=-4

. A length cannot be negative, so

x=6

is the meaningful value.

Question 3

A student solves a quadratic model and gets two positive values. What should the student do next?

Choose the larger value without checking.
Check both values in the original equation and compare both with the situation.
Add the two values and report the sum.
Reject both values because a quadratic must have only one solution.

Show answer and explanation

Check both values in the original equation and compare both with the situation.

Both values may work algebraically. Substitute each into the original relationship and use the story to decide which answer or answers apply.

Key terms

Variable: A letter that represents an unknown or changing number.
Quadratic equation: An equation that includes a squared variable and can be written with a term such as $x^2$ .
Model: A mathematical equation or representation of a situation.
Factor: One of the expressions multiplied together to form a product.
Solution: A value that makes an equation true.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCF3M), expectation A3.3. It is a study resource, not an official curriculum publication.

Official curriculum reference

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A3.3 · Solve real-world problems from quadratic equations

Ontario Grade 11 Mathematics

Turn a situation into an equation, solve it, and decide what the solutions mean.

What you will learn

1. Prerequisite bridge: quantities and equations

2. From a situation to a quadratic model

3. Solve, check, and interpret

4. A reliable problem-solving routine

Turning the patio information into mathematics

Finding the dimensions of a patio

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MCF3M

About this lesson