A1.9 Solve applied quadratic problems by factoring or graphing

Ontario Grade 11 Mathematics

Mathematical Models

Ontario Grade 11 MBF3C — study-guide label A1.9

A quadratic model describes a situation where a changing quantity is related to the square of another quantity. It can describe, for example, the area of a rectangle whose sides change together or the height of an object over time. In an applied problem, solving the equation is only part of the job: you must also decide what the solutions mean and whether they make sense. This lesson uses two Grade 11 methods: factoring and graphing. Before either method, recall that a solution to an equation is a value that makes both sides equal. Also remember that measurements such as length and time cannot be negative in many real situations, even if an equation produces a negative number.

What you will learn

Recognize when a practical situation can be represented by a quadratic equation or graph.
Use factoring to find possible solutions and decide which solution fits the situation.
Use a graph to estimate or identify when a quantity reaches a stated value.
State answers with suitable units and explain what they mean in context.

1. From a situation to a quadratic equation

A variable is a letter used to stand for an unknown or changing number. A quadratic expression has a squared variable as its highest power, such as

x^2+3x-40

. A quadratic equation sets a quadratic expression equal to another quantity, such as

x^2+3x-40=0

.

Applied means connected to a real situation. Start by deciding what the variable represents and recording its unit. Then translate the relationships in the problem into an equation or graph. For instance, if a rectangle has width

x

metres and length

x+3

metres, its area is

x(x+3)

square metres. The product creates a squared term when expanded.

Factoring rewrites an expression as a product of simpler expressions. It is useful when setting the model equal to a target value leads to factors that can be identified. The zero-product property says that if two factors multiply to zero, at least one factor must be zero. This lets us find the values that make a factored quadratic equation true.

A graph shows how the output of a model changes as its input changes. To solve a problem by graphing, look for where the model reaches the target value. If the target is zero, these points are called the zeros or x-intercepts. If the target is not zero, graph the model and the target level, then read their intersection points. A graph may give an estimate, so use an appropriate scale and report the precision shown.

Name the variable and its unit before building the model.
An equation's solutions must be checked against the situation.
Factoring finds solutions through products equal to zero; graphing finds inputs that produce a target output.

2. Choosing a method and judging solutions

Use factoring when the equation can be rearranged into a product equal to zero and its factors are manageable. For example, an expression such as

x^2+3x-40

can be written as

(x+8)(x-5)

. Each factor gives a possible solution when set equal to zero.

Use graphing when the model and target can be displayed clearly and the intersections can be read. A graph is especially helpful when the problem asks when a quantity reaches a particular level. You can use graphing technology or a carefully drawn graph, but the axes, scale, and labels must match the quantities and units.

A quadratic equation can have two solutions, one solution, or no real solutions. In an applied setting, even two mathematical solutions may not both be meaningful. A negative width, for example, does not describe a physical rectangle. A negative time may be outside the time period being studied. Keep only solutions that fit the stated situation, and explain why.

Check a candidate solution by substituting it into the original relationship or by checking that the graph reaches the target there. Also check units: area is measured in square units, while a side length is measured in units of length. Do not report an area as though it were a length.

Set the equation equal to zero before using the zero-product property.
Graph intersections answer the question of where two quantities have the same value.
Reject values that conflict with the situation, but state why they are rejected.

3. Reading and communicating an applied answer

A good solution connects the calculation to the question. Write what the answer measures, include units when relevant, and distinguish an exact value from a graph estimate. If graphing technology displays an intersection at about

t=1.0

, report it as an approximate time rather than claiming greater precision than the graph supports.

Before finishing, ask three questions: Does the value satisfy the equation or graph? Is it possible in the situation? Does the answer use the requested quantity and unit? These checks help prevent a mathematically correct number from becoming an incorrect applied answer.

The two examples that follow use different models and different methods. The first uses area and factoring. The second uses height and graphing to find the times when a target height is reached.

Use words and units to make the meaning of a solution clear.
Match the precision of a graph-based answer to the graph or technology display.
Check both the mathematics and the context.

Garden model: what each quantity represents

Quantity	Expression	Unit
Width	$x$	m
Length	$x+3$	m
Area	$x(x+3)$	m²
Given area	$40$	m²

Worked example

Finding the dimensions of a garden

A rectangular garden has a length that is 3 m more than its width. Its area is 40 square metres. Find its width and length.

Choose a variable
Let $x$ be the garden's width in metres. Since the length is 3 m more, its length is $x+3$ metres. Both dimensions must be positive.
Build the area equation
Area equals width times length. The product of the two side lengths must equal the given area of 40 square metres.
$x(x+3)=40$
Write the equation in factored form
Move 40 to the left so the equation equals zero. The expression factors because the numbers 8 and negative 5 multiply to negative 40 and add to 3.
$(x+8)(x-5)=0$
Find and interpret the solutions
The zero-product property gives the possible widths $x=-8$ or $x=5$ . A negative width is impossible, so use $x=5$ m. The length is $5+3=8$ m.

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

Check: The area is

5\times 8=40

square metres, and the length is 3 m more than the width. Both conditions are satisfied.

Worked example

Finding when a height reaches a target

A model for the height of a ball is

h=-5t^2+20t+1

, where

h

is height in metres and

t

is time in seconds after the ball is released. Use a graph of the model to find when the ball is at a height of 16 m.

Set the target
The question asks when the model's height equals 16 m. Graph the height model and the horizontal target level $h=16$ . Their intersection points show the times when both have the same height.
Read the intersections
On a graph with time on the horizontal axis and height on the vertical axis, the model meets the 16 m level at about $t=1$ s and $t=3$ s. These are the two times to report; the graph shows that the target height is reached once on the way up and once on the way down.
Check the read values
Substitution confirms both readings: at 1 s the model gives 16 m, and at 3 s it also gives 16 m. The times are non-negative, so both fit the stated time period.
$-5(1)^2+20(1)+1=16,\quad -5(3)^2+20(3)+1=16$

Answer: The ball is at 16 m at about 1 s and 3 s after release.

Check: Both times produce a height of 16 m in the given model. The graph's two intersections match the substitution check.

Common mistakes and how to avoid them

Keeping every solution from a factored equation.

Correction: Check each value in the situation. A negative width cannot describe the garden, so reject it and explain why.

Reading the height value as the time on a graph.

Correction: Use the horizontal axis for time and the vertical axis for height. The intersection's horizontal coordinate gives the time.

Giving a graph estimate as an exact answer.

Correction: Use words such as “about” when the graph only supports an estimate, and do not claim more decimal places than are shown.

Forgetting units or using the wrong kind of unit.

Correction: State side lengths in metres, time in seconds, and area in square metres.

Lesson summary

Translate the situation into a quadratic equation or graph, and identify what the variable represents.
Solve by factoring or graphing: factoring uses products equal to zero, while graphing uses target-level intersections.
Check the result in the model, reject values that do not fit the context, and state the answer with suitable units and precision.

Check your understanding

Question 1

A rectangular pen has width

x

metres and length

x+2

metres. Its area is 48 square metres. Which equation represents the situation?

$x(x+2)=48$
$x+(x+2)=48$
$2x+2=48$
$x^2+2=48$

Show answer and explanation

x(x+2)=48

Area is found by multiplying width by length. The dimensions are

x

and

x+2

, so their product equals 48.

Question 2

A graph of a height model meets the target level at

t=2

and

t=5

. What do these two values represent?

The two heights of the object
The two times the object is at the target height
The object's starting and ending heights
The height and time units

Show answer and explanation

The two times the object is at the target height

The horizontal coordinate on this graph is time. Each intersection gives a time when the model's height equals the target.

Question 3

A factored area equation gives possible widths of

-6

m and

4

m. Which width should be used for a real rectangle?

$-6$ m, because it is the smaller solution
$4$ m, because a width must be positive
Both values, because both solve the equation
Neither value, because an area cannot have solutions

Show answer and explanation

4

m, because a width must be positive

The negative value may satisfy the equation, but a real rectangle cannot have a negative width. The positive width is the meaningful solution.

Key terms

Quadratic expression: An expression whose highest variable power is 2, such as $x^2+3x-40$ .
Factoring: Rewriting an expression as a product of simpler expressions.
Zero-product property: If a product equals zero, at least one of its factors must equal zero.
Intersection: A point where two graphs meet; its coordinates give values shared by both graphs.
Solution: A value that makes an equation true or identifies a point on a graph that meets the problem's target.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MBF3C), expectation A1.9. It is a study resource, not an official curriculum publication.

Official curriculum reference

Report a correction or ask a question

A1.9 · Solve applied quadratic problems by factoring or graphing

Ontario Grade 11 Mathematics

Ontario Grade 11 MBF3C — study-guide label A1.9

What you will learn

1. From a situation to a quadratic equation

2. Choosing a method and judging solutions

3. Reading and communicating an applied answer

Garden model: what each quantity represents

Finding the dimensions of a garden

Finding when a height reaches a target

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MBF3C

About this lesson