Chapter 3 – Solving Quadratics Word Problems Where You Have to Create the Equation

Agenda

• Movement problems (creating the equation)*• Engineering problems (creating the equation)*• Revenue problems (creating the equation)*• Area problems (creating the equation)*• Integer problems (creating the equation)• Triangle problems (creating the equation)• * Indicates done within this lecture

Mental Health Break

Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14

• Angus is playing golf. The diagram (not to scale, shows him making a perfect shot to the hole. Determine the height of the ball when it is 15m from the hole by using the information in the diagram to determine a quadratic relation for height v. distance travelled

• To find the height we will need the equation• To find the equation we need the zeros and

one other point• Zeros… • x of vertex…• y of vertex…

one other point• Zeros are 0 and 100 • x of vertex = (0+100)/2 = 50• y of vertex… 10 (height of the tree in the

middle)

• Use y = a(x – S)(x – t) • put in the zeros of 0 and 100 for S & t*• put in the point (50, 10) for x & y

• *Note – you would end up with the same “a” if you used -50 and 50 for the zeros and made x = 0 the centre of the graph

• y = a(x – S)(x – t) • 10 = a(50 – 0)(50 – 100) 10 = a(50)(-50) 10 = a(-2500) 10/-2500 = a(-2500)/-2500• a = -0.004• y = -0.004x(x – 100)• y = -0.004x²+0.4x

• y = -0.004x²+0.4x• We shouldn’t cannot plug in an x of 15m

because we are 15m from where the ball lands• If we did, we could argue symmetry however• Ideally, we plug in 85m for x because 100m –

15m = 85m and we can use that distance to then calculate the height of the ball at that time

• y = -0.004x² + 0.4x• If we plug in x of 85m, we can calculate y (the

height of the ball at that time)• y = -0.004(85)² + 0.4(85)• y = -0.004(7225) + 34• y = -28.9 + 34• y = 5.1m• So, the height of the ball 15m from the hole is

Quadratic Word Problems Engineering Creating the Equation – Ex. P.284, #16

• The second span of the Bluewater Bridge, in Sarnia, Ontario, is supported by a pair of steel parabolic arches. The arches are set in concrete foundations that are on opposite sides of the St. Clair River 281m apart. The top of each arch rises 71m above the river. Determine the algebraic expression that models the arch and find the height of the span 10 metres from the left side of the river.

one other point• Zeros… • x of vertex…• y of vertex…

one other point• Zeros are 0 and 281 • x of vertex = (0+281)/2 = 140.5• y of vertex… 71 (height of the span of bridge in

the middle)

• Use y = a(x – S)(x – t) • put in the zeros of 0 and 281 for S & t*• put in the point (140.5, 71) for x & y

• *Note – you would end up with the same “a” if you used –140.5 and 140.5 for the zeros and made x = 0 the centre of the graph

• y = a(x – S)(x – t) • 71 = a(140.5 – 0)(140.5 – 281) 71 = a(140.5)(-140.5) 71 = a(-19740.25) 71/-19740.25 = a(71)/-19740.25• a = -0.004• y = -0.004(x - 0)(x – 281)• y = -0.004x(x – 281)• y = -0.004x² + 1.124x

• y = -0.004x²+1.124x• We can plug in an x of 10m

• y = -0.004x² + 1.124x• If we plug in x of 10m, we can calculate y (the

height of the ball at that time)• y = -0.004(10)² + 1.124(10)• y = -0.004(100) + 11.24• y = -0.4 + 11.24 • y = 10.84• So, the height of the bridge span 10m from the

left side of the river is 10.84m

Quadratic Word Problems Revenue Creating the Equation

• Recall, 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

• Icky Bum Ltd. research shows that a $0.50 increase in the price of a box of baby wipes results in 5 fewer boxes of baby wipes being sold. The usual price of $20 for a box of baby wipes results in a sale of 280 boxes

• What should the Icky Bum do? Should they “stick” with the current price?

• Equation set-up is always the same for this type of Revenue problem

• R = (orig. price + ∆ in price) (orig. # sold + ∆ in number sold)

• R = (20 + 0.50x)(280 – 5x)

• R = (orig. price + ∆ in price) (orig. # sold + ∆ in number sold)

• R = (20 + 0.50x)(280 – 5x)• Our strategy is going to be to set-up the

equation, find the zeros, take the average of the zeros to find the x of the vertex, and substitute x back in and solve for the vertex R which will give us the maximum revenue

• R = (20 + 0.50x)(280 – 5x)• 0 = (20 + 0.50x)(280 – 5x)• 20 + 0.50x = 0 and 280 – 5x = 0 (set each bracket = 0)• 0.50x = -20 and -5x = -280• x = -40 and x = 56

• R = (20 + 0.50x)(280 – 5x)• Zeros are -40 and 56• xv = xzero 1 + xzero 2

---------------------------

2• xv = (-40 + 56) --------................ xv = 8 2

• R = (20 + 0.50x)(280 – 5x)• xv = 8• R = (20 + 0.50(8))(280 – 5(8))• R = (24)(240)• So, maximum revenue is $5,760 (note –

previously, it was $20 x 280 = $5,600)

Quadratic Word Problems Area Creating the Equation

• When creating quadratic area equations, you are quite often working with either the area equation alone, or with the perimeter equation as well.

• When working with maximum area created by enclosure of so much material, you will need both equations

• When you are working with a maximum area of some type of border, you are usually working only with the area equation

• A = l x w (area)• P = 2l + 2w (perimeter)

• Ex. Say you want to find equation showing the maximum enclosure area given a perimeter of 20m

• P = 2l + 2w• 20 = 2l + 2w (substitute 20 in for P)• 10 = l + w (divide both sides by 2)• l = 10 – w (isolate l)

• Now, substitute l = 10 – w into A = l x w• A = (10 – w)(w) (factored form)• A = 10w - w² (standard form)• A = -w² + 10w ( standard form)

• A = -w² + 10w • 0 = - w² + 10w• 0 = -w(w + 10)• -w = 0 and w + 10 = 0• Zeros are 0 and -10• wv = -5 (take average of zeros)• A = -(5)²+10(5) = 25• So, a width of 5m results in the maximum area of

• Mr. and Mrs. B want to install a rectangular swimming pool measuring 10m x 20m. They want to put a deck of uniform width around the pool. They have budgeted $1,920 to spend on the deck and know that construction and material costs are $30/metre

• Here we are using the area equation only, but we have to consider the two areas – that of the pool & that of the deck surrounding it

• Let’s let x represent the width of the uniform deck surrounding the pool

• Then, let’s state what the “new” overall length and width are of the pool & the deck together

• We know the inside area is:• A = 10 x 20 = 200m²• For the outside area we have to use other

information given to determine it. We know that the B.’s have budgeted $1,920 to spend on the deck and that construction & material costs are $30 per metre.

• How can we get an area out of this?

• We know that the B.’s have budgeted $1,920 to spend on the deck and that construction & material costs are $30 per metre.

• How can we get an area out of this?• Divide $1920 by $30 = 64• So, the total area of the pool and the deck will

be 264 m²

• So, what is our equation?• A = l x w• 264 = (20 + 2x)(10 + 2x)• 264 = 200 + 60x + 4x²• 0 = 4x² + 60x + 200 – 264• 0 = 4x² + 60x – 64• 0 = 4(x² + 15x – 16)• 0 = 4(x + 16)(x – 1)• x = -16 or x = 1 (reject -16) so, the width of the deck

should be 1m

Quadratic Word Problems Integers Creating the Problem

• The key with integer word problems is the translation of words into algebra

• Consecutive numbers

• For consecutive numbers, you have to define x as any integer and the next number would be x + 1 (ex. If x was one, x + 1 = 2)

• Consecutive odd numbers

• For consecutive odd numbers, you have to define x as any odd integer and the next consecutive odd number would be x +2 (ex. If x was one, x + 2 = 3)

• Consecutive even numbers

• For consecutive even numbers, you have to define x as any even integer and the next consecutive even number would be x +2 (ex. If x was -4, x + 2 = -2)

• Sum of squares

• Means you have to “square” or multiply each term by itself and then add the sum of the squares

• Ex. The sum of the squares of 2 consecutive numbers…

• Sum of squares• Ex. The sum of the squares of 2 consecutive numbers… x and x +

1• Sum of squares would be:= (x)²+ (x + 1)² = x²+ (x² + x + x + 1) (square first bracket and apply foil to the second

bracket)= 2x²+ 2x + 1• Typically then this would be set to a number and you have to

bring that number to the left, subtract it from 1 and create a trinomial (which can be factored)

• Let’s do a full example now…

• Ex. The sum of the squares of 2 consecutive odd integers is 34. Find the two integers

• Let x be the 1st odd integer• So, x + 2 would be the next odd integer• Sum of the squares means we have to have an

(x)² and an (x + 2)²• But, we are told the sum of these has to = 34• (x)² + (x + 2)² = 34

• (x)² + (x + 2)² = 34• x² + x² + 2x + 2x + 4 = 34• 2 x² + 4x + 4 = 34• 2 x² + 4x + 4 – 34 = 0• 2 x² + 4x – 30 = 0• 2( x² + 2x – 15) = 0• 2 (x + 5)(x – 3) = 0• So, the two zeros are -5 and 3 (these are also the

two possible integers that work)

Homework (Given the Equation)

• Handout

Chapter 3 – Solving Quadratics Word Problems Where You Have to Create the Equation

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