A. Quadratic Functions · Quadratic Expressions and Equations

Solving Problems with Quadratic Equations

A1.1 A1.2 A1.4 A1.6 A1.8About 75 minutes

Learning goals

I can write a quadratic equation to represent a real situation.
I can choose an efficient strategy and solve the equation.
I can interpret the roots in context, reject impossible ones, and check my answer.

Words to know

uniform width
The same width all the way around, as for a path or frame.

Pythagorean theorem
In a right triangle with legs a and b and hypotenuse c, a² + b² = c².

interpret
Explain what a mathematical answer means in the real situation.

reject (a root)
Decide that a correct root of the equation is not a valid answer to the problem, and say why.

Putting it all together

This lesson combines everything in the unit: you will represent a situation with a quadratic expression, write an equation, choose a strategy to solve it, and interpret the roots. Real problems rarely say "factor this". They describe a garden, a price, a punt or a ladder, and you decide what to do.

A problem-solving plan

Understand. What is being asked? Draw a diagram or make a small table.
Plan. Define the variable with units. Write the equation.
Solve. Rearrange to ax² + bx + c = 0. Check the discriminant if you are not sure how many roots to expect. Choose factoring, the formula or graphing.
Check and conclude. Reject roots that do not make sense. Check the answer in the original situation. Answer in a sentence with units.

A 12 m by 8 m garden with a path of width x all around. The outer rectangle is (12 + 2x) by (8 + 2x). The picture is drawn to scale using the answer, x = 1.5 m.

Worked example

Example 1: A garden path

A rectangular garden 12 m by 8 m is surrounded by a path of uniform width. The garden and path together cover 165 m². How wide is the path?

Let x be the path width in metres. The outer rectangle is (12 + 2x) by (8 + 2x).
Equation: (12 + 2x)(8 + 2x) = 165. Expand: 96 + 24x + 16x + 4x² = 165, so 4x² + 40x − 69 = 0.
Discriminant: 1600 − 4(4)(−69) = 1600 + 1104 = 2704 = 52². A perfect square, so the roots are rational.
Formula: x = (−40 ± 52) / 8, so x = 12/8 = 1.5 or x = −92/8 = −11.5. A width cannot be negative: reject −11.5.
Check: the outer rectangle is 15 m by 11 m, and 15 × 11 = 165 ✓.

The path is 1.5 m wide.

Worked example

Example 2: Pricing butter tarts

A bakery sells 200 butter tarts a day at $2.00 each. For every $0.10 price increase, it sells 5 fewer tarts. What price(s) give a daily revenue of $442?

Let x be the number of $0.10 increases. Price = 2 + 0.1x; number sold = 200 − 5x.
R = (2 + 0.1x)(200 − 5x) = 400 − 10x + 20x − 0.5x² = −0.5x² + 10x + 400.
Set R = 442: −0.5x² + 10x − 42 = 0. Multiply by −2: x² − 20x + 84 = 0.
Factor: product 84, sum −20: −6 and −14. (x − 6)(x − 14) = 0, so x = 6 or x = 14.
x = 6: price $2.60, 170 sold, 2.60 × 170 = $442 ✓. x = 14: price $3.40, 130 sold, 3.40 × 130 = $442 ✓.

Either $2.60 or $3.40 gives $442. (The bakery might prefer $2.60: more customers get tarts for the same revenue.)

Worked example

Example 3: A punted football

A football is punted so that its height in metres after t seconds is h = −4.9t² + 18t + 1. For how long is the ball higher than 15 m?

Find when h = 15: −4.9t² + 18t + 1 = 15, so −4.9t² + 18t − 14 = 0.
Decimals, so use the formula. D = 324 − 4(−4.9)(−14) = 324 − 274.4 = 49.6. √49.6 ≈ 7.043.
t = (−18 ± 7.043) / (−9.8): t ≈ 1.12 or t ≈ 2.56.
The parabola opens down, so the ball is above 15 m between these times: 2.56 − 1.12 ≈ 1.44 s.

The ball is above 15 m for about 1.44 s (from about 1.12 s to 2.56 s).

The football's height. The shaded band shows the times when the ball is above the dashed line h = 15 m: between about 1.12 s and 2.56 s.

Worked example

Example 4: A right triangle

One leg of a right triangle is 7 cm longer than the other, and the hypotenuse is 13 cm. Find the lengths of the legs.

Let the shorter leg be x cm; the longer leg is x + 7.
Pythagorean theorem: x² + (x + 7)² = 13². Expand: x² + x² + 14x + 49 = 169, so 2x² + 14x − 120 = 0.
Divide by 2: x² + 7x − 60 = 0. Factor: (x + 12)(x − 5) = 0.
x = 5 or x = −12. Reject the negative length. The legs are 5 cm and 12 cm.
Check: 5² + 12² = 25 + 144 = 169 = 13² ✓.

The legs are 5 cm and 12 cm.

The triangle from Example 4. Using the Pythagorean theorem with x and x + 7 creates a quadratic equation.

Checklist for word problems

Did you define the variable, with units?
Did you write the equation from the situation (not from guessing)?
Did you reject impossible roots and say why?
Did you check in the original words (not just the equation)?
Did you answer the question asked (a price, not just x)?

Just for fun

Butter tarts are proudly Canadian, and they even help with algebra: if a pan makes 12 and you eat 3 "just to test them", the remaining tarts are a very real restriction on the domain.

Hands-on activity

Design a border for a real object

You need: a measuring tape, a calculator, and a rectangular object such as a placemat, a cutting board or a poster.

Measure your rectangle's length L and width W (in cm).
You want to add a border of equal width x all around so that the total area is exactly double the original area.
Write the equation (L + 2x)(W + 2x) = 2LW, rearrange it to ax² + bx + c = 0, and solve it with the quadratic formula.
Reject any negative root. Mark the border on paper around your object to see whether the answer looks right.

What to look for: for a 40 cm by 30 cm placemat, (40 + 2x)(30 + 2x) = 2400 gives 4x² + 140x − 1200 = 0, or x² + 35x − 300 = 0. D = 1225 + 1200 = 2425, so x = (−35 + √2425)/2 ≈ 7.12 cm. A border only about 7 cm wide doubles the area, which surprises most people.

Your turn

The product of two consecutive integers is 156. Find the integers.

A rectangle has perimeter 34 cm and area 60 cm². Find its dimensions.

A ball's height is h = −5t² + 20t + 2. When is it 17 m high?

A 20 cm by 15 cm photo is framed so that the total area is 374 cm². How wide is the frame?

A right triangle has hypotenuse 10 m, and one leg is 2 m longer than the other. Find the legs.

The theatre revenue from Representing Situations with Quadratic Expressions is R = −20x² + 60x + 3600, where x is the number of $1 increases on a $12 ticket. What ticket prices give $3640?

A model rocket's height is h = −4.9t² + 30t. When does it land?

More practice: warm-up

A number squared plus 3 times the number is 40. Find the number(s).

A square has area 49 m². How long is a side?

A ball's height is h = −5t² + 10t. When is it back on the ground?

A rectangle is twice as long as it is wide and has area 50 cm². Find its dimensions.

Why do we reject a negative root when the variable is time after a throw?

More practice: on your own

A 10 m by 6 m garden has a path of uniform width; together they cover 96 m². How wide is the path?

When does the punted football in Example 3 (h = −4.9t² + 18t + 1) land?

For the butter tarts, R = −0.5x² + 10x + 400. Besides $2.00, what price gives revenue of $400?

A picture has area 300 cm², and its length is 5 cm more than its width. Find its dimensions.

A 5 m ladder leans against a wall. The top is 1 m higher up the wall than the foot is from the wall. How far is the foot from the wall?

Challenge

A 30 cm wide sheet is folded into a gutter with sides x cm high (cross-section area x(30 − 2x)). Find all fold heights that give an area of 100 cm².

The canoe outfitter's revenue is R = −10x² + 90x + 3600 for x increases of $2 on a $30 rental. What rental price(s) give $3800? Which would you recommend?

A diver jumps from a 10 m platform: h = −4.9t² + 3t + 10. (a) When does she reach the water? (b) Does she ever reach 11 m above the water?

Check your understanding

Quiz

A 6 m by 4 m patio gets a border of width x. Which equation says the total area is 48 m²?

Solving a length problem gives x = 3 or x = −9. The length is:

Legs x and x + 1 and hypotenuse 5 lead to which equation?

A height equation h = 20 has roots t = 0.8 and t = 3.2. The object is above 20 m:

What is the last step of the problem-solving plan?

Revenue R = −0.5x² + 10x + 400 equals $442 at x = 6 or 14. With price 2 + 0.1x, the prices are:

For parents and teachers

Big idea: solving real problems with quadratic equations: representing the situation (A1.2), choosing and applying a solving strategy (A1.4, A1.6, A1.8) and interpreting the results using the graph (A1.1).

How to help: read a problem together and ask your teen to explain the plan before doing any algebra: What is x? What equation? Which method? Which answer will we reject? Insist on a final sentence that answers the question with units.

Common difficulties:

Writing the equation is the hard part. A labelled diagram or a small table (x = 0, 1, 2) helps.
Stopping at x instead of answering the question (a price, both dimensions, a length of time).
Forgetting the "between the roots" idea for questions like "for how long is it above 15 m?"
Not checking the answer in the original words of the problem.