# Solving Word Problems Given the Quadratic Equation

## Presentation on theme: "Solving Word Problems Given the Quadratic Equation"— Presentation transcript:

Solving Word Problems Given the Quadratic Equation

Learning Goal By the end of the lesson, students will be able to… apply knowledge of quadratics to solving word problems

Curriculum Expectations
By the end of the lesson, students will… Solve problems arising from realistic situations represented by a graph or an equation of a quadratic relationship

Mathematical Process Expectations
Connecting – make connections among mathematical concepts and procedures; and relate mathematical ideas to situations or phenomena drawn from other contexts

Agenda Solving quadratic word problems generally
Movement problems (given the equation) Engineering problems (given the equation) Revenue problems (given the equation) Area problems (given the equation)

Mental Health Break

For quadratic word problems, you will either be given, or need to set-up a quadratic equation in either standard or factored form If the problem asks when something happens, you must find the x value of the vertex, or sometimes the zeros (ex. When ball hits ground) If the problem asks for the maximum, minimum or optimal value, you must find the y value of the vertex

Problems usually fall into one of the following categories Movement (ex. Ball thrown) Engineering (ex. Bridge) Revenue (ex. Price change or maximum revenue) Area (ex. Maximum area or how much border can you add) Integers (ex. Squares of consecutive odd integers) Triangles (ex. Find side length of right angle triangle)

Solving Quadratic Word Problems Given the Equation
Today, we’re going to look at some problems where you are given the equation Movement (ex. Ball thrown) Revenue (ex. Break-even and maximum revenue) Area (ex. Maximum area)

Quadratic Word Problems Movement Given the Equation
Factor (GCF) if possible Set y = 0 and determine the zeros Answer questions about time (x-axis) or height (y-axis) Typically have to determine the vertex Remember, average of zeros gives x of vertex Then, substitute x into equation and solve for y of vertex

Quadratic Word Problems Movement Given the Equation
If asked for the x value for a given y value other than zero, substitute it in for y but then move it to the other side so that there is a zero where the y was again You then should end up with a trinomial that you can factor and solve for x again These won’t be the same x’s as the zeros

Quadratic Word Problems Movement Given the Equation, ex. p.268, #11
A football is kicked straight up into the air. Its height above the ground is approximated by the relation h = 25t – 5t², where h is the height in metres and t is the time in seconds

Quadratic Word Problems Movement Given the Equation, ex. p.268, #11

Quadratic Word Problems Movement Given the Equation, ex. p.268, #11
h = 25t – 5t² (a) What are the zeros of the relation and when does the football hit the ground? 0 = 25t – 5t² Set h = 0 0 = 5t(5 – t) Factor our the GCF 5t = 0 and 5 – t = Isolate mini-zero equations t = 0 and t = These are the two zeros

Quadratic Word Problems Movement Given the Equation, ex. p.268, #11
h = 25t – 5t² Since the zeros are 0 and 5, the football hits the ground at t = 5 or 5 seconds

Quadratic Word Problems Movement Given the Equation, ex. p.268, #11
h = 25t – 5t² (b) What are the coordinates of the vertex Zeros are 0 and 5 tv = tzero 1 + tzero 2 2 tv = (0 + 5) tv = 2.5

Quadratic Word Problems Movement Given the Equation, ex. p.268, #11
h = 25t – 5t² (b) What are the coordinates of the vertex t of vertex is 2.5. Substitute this value in for t and solve for h h = 25(2.5) – 5(2.5) ² h = 62.5 – 31.25 h = 31.25m (d) The maximum height of 31.25m occurs at time of 2.5 seconds

Quadratic Word Problems Movement Given the Equation, ex. p
Quadratic Word Problems Movement Given the Equation, ex. p.268, #11(c) (Graph)

Quadratic Word Problems Revenue Given the Equation
Equation given is usually either a Revenue Equation or a Profit Equation Revenue basically means money from sales and is calculated by taking the price of a item and multiplying it by the number of items sold R = price x # sold Profit is the revenue of a company after expenses are subtracted Breakeven takes place when profit (not revenue) is zero Maximum profit or revenue takes place at the vertex of the graph

Quadratic Word Problems Revenue Given the Equation, ex. p.332, #15
The Wheely Fast Co. makes custom skateboards for professional riders. They model their profit with the relation P = -2b² + 14b – 20, where b is the number of skateboards they produce (in thousands), and P is the company’s profit in hundreds of thousands of dollars

Quadratic Word Problems Revenue Given the Equation, ex. p.332, #15

Quadratic Word Problems Revenue Given the Equation, ex. p.332, #15
P = -2b² + 14b – 20 When does Wheely Fast break even (P = 0) 0 = -2b² + 14b – set P = 0 0 = -2(b² + 7b – 10) factor out -2, GCF 0 = -2(b – 5)(b – 2) factor using butterfly b – 5 = 0 and b – 2 = Identify zero equations b = 5 and b = The zeros are 5 and 2 so this happens when they produce either 2000 or 5000 skateboards

Quadratic Word Problems Revenue Given the Equation, ex. p.332, #15

Quadratic Word Problems Revenue Given the Equation, ex. p.332, #15
P = -2b² + 14b – 20 (b) How many skateboards does Wheely Fast need to produce to maximize profit Zeros are 2 and 5 bv = bzero 1 + bzero 2 2 bv = (2 + 5) bv = 3.5 (which is 3500)

Quadratic Word Problems Revenue Given the Equation, ex. p.332, #15
P = -2b² + 14b – 20 Substitute b = 3.5 into equation P = -2(3.5)² + 14(3.5) – 20 P = -2(12.25) + 49 – 20 P = – 20 P = 4.5 (so 4.5 x 100,000 = \$450,000) So, the Wheely Fast maximizes profits of \$450,000 by producing 3,500 skateboards

Quadratic Word Problems Area Given the Equation, ex. p.308, #9
A rectangular enclosure has an area in square metres given by A = -2w² + 36w, where w is the width of the rectangle in metres. What is the maximum area of the enclosure?

Quadratic Word Problems Area Given the Equation, ex. p.308, #9

Quadratic Word Problems Area Given the Equation, ex. p.308, #9
A = -2w² + 36w First, find the zeros 0 = -2w² + 36w Set A = 0 0 = -2w(w - 18) Factor our the GCF -2w = 0 and w – 18 = Isolate mini-zero equations w = 0 and w = These are the two zeros

Quadratic Word Problems Area Given the Equation, ex. p.308, #9
A = -2w² + 36w What are the coordinates of the vertex Zeros are 0 and 18 wv = wzero 1 + wzero 2 2 wv = (0 + 18) wv = 9

Quadratic Word Problems Area Given the Equation, ex. p.308, #9
Now we need to find the A of the vertex. The w of the vertex is 9. Substitute this value in for w and solve for A A = -2w² + 36w A = -2(9)²+ 36(9) A = -2(81) + 324 A = 162 The maximum height of area of 162m² occurs when w is 9 metres

Quadratic Word Problems Area Given the Equation, ex. p.308, #9
Maximum area (vertex) Zero Zero

Homework (Given the Equation)
Handout 1 – Quadratic Word Problems - #3 & 4 Handout 2 – Application Problems - #1, 5 & 6