by Mr Johari
source: http://www.mathsisfun.com/algebra/quadraticequationrealworld.html
REAL LIFE APPLICATION OF QUADRATIC EQUATION
EXAMPLE 1
 
Balls, Arrows, Missiles and Stones
If you throw a ball (or shoot an arrow, fire a missile or throw a stone) it will go up into the air, slowing down as it goes, then come down again ...
Example: Throwing a Ball
A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. When does it hit the ground?

Ignoring air resistance, we can work out its height by adding up these three things:
The height starts at 3 m:   3 
It travels upwards at 14 meters per second (14 m/s):   14t 
Gravity pulls it down, changing its speed by about 5 m/s per second (5 m/s^{2}):   5t^{2} 
(Note for the enthusiastic: the 5t^{2} comes from ½at^{2} with a=9.81 m/s^{2})   
Add them up and the height h at any time t is:
h = 3 + 14t  5t^{2}
And the ball will hit the ground when the height is zero:
3 + 14t  5t^{2} = 0
5t^{2} + 14t + 3 = 0
Let us solve it ...
There are many ways to solve it, here we will use the
factoring method:
Will be easier if we multiply all terms by 1:   5t^{2}  14t  3 = 0 
Now our job is to factor it. We will use the "Find two numbers that multiply to give a×c, and add to give b" method in Factoring Quadratics.
a×c = 15, and b = 14.
The positive factors of 15 are 1, 3, 5, 15, and one of the factors has to be negative. By trying a few we find that 15 and 1 work (15×1=15, and 15+1= 14)

Rewrite middle with 15 and 1:   5t^{2}  15t + t  3= 0 
Factor first two and last two:   5t(t  3) + 1(t  3) = 0 
Common Factor is (t  3):   (5t + 1)(t  3) = 0 
  
And the two solutions are:   5t + 1 = 0 or t  3 = 0 
  t = 0.2 or t = 3 
The "t = 0.2" is a negative time, impossible in our case.
The "t = 3" is the answer we want:
The ball hits the ground after 3 seconds!

Here is the graph of the Parabola h = 5t^{2} + 14t + 3
It shows you the height of the ball vs time
Some interesting points:
(0,3) When t=0 (at the start) the ball is at 3 m
(0.2,0) Says that 0.2 seconds BEFORE we threw the ball it was at ground level ... this never happened, so our common sense says to ignore it!
(3,0) Says that at 3 seconds the ball is at ground level.
Note also that the ball reaches nearly 13 meters high.
 
Note for the enthusiastic: You can find exactly where the top point is! The method is explained in
Graphing Quadratic Equations, and has two steps:
Find where (along the horizontal axis) the top occurs using b/2a:
 t = b/2a = (14)/(2 × 5) = 14/10 = 1.4 seconds
Then find the height using that value (1.4)
 h = 5t^{2} + 14t + 3 = 5(1.4)^{2} + 14 × 1.4 + 3 = 12.8 meters
So the ball reaches the highest point of 12.8 meters after 1.4 seconds.
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EXAMPLE 2
Example: New Sports Bike
 
You have designed a new style of sports bicycle!
Now you want to make lots of them and sell them for profit.

Your costs are going to be:
 $700,000 for manufacturing setup costs, advertising, etc
 $110 to make each bike
Based on similar bikes, you can expect sales to follow this "Demand Curve":
 Unit Sales = 70,000  200P
Where "P" is the price.
For example, if you set the price:
 at $0, you would just give away 70,000 bikes
 at $350, you would not sell any bikes at all.
 at $300 you might sell 70,000  200×300 = 10,000 bikes
  
So ... what is the best price? And how many should you make?
Let us make some equations!
How many you sell depends on price, so use "P" for Price as the variable
 Unit Sales = 70,000  200P
 Sales in Dollars = Units × Price = (70,000  200P) × P = 70,000P  200P^{2}
 Costs = 700,000 + 110 x (70,000  200P) = 700,000 + 7,700,000  22,000P = 8,400,000  22,000P
 Profit = SalesCosts = 70,000P  200P^{2}  (8,400,000  22,000P) = 200P^{2} + 92,000P  8,400,000
Profit = 200P^{2} + 92,000P  8,400,000
Solve: 200P^{2} + 92,000P  8,400,000 = 0
Step 1 Divide all terms by 200
P^{2} – 460P + 42000 = 0
Step 2 Move the number term to the right side of the equation:
P^{2} – 460P = 42000
Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:
(b/2)^{2} = (460/2)^{2} = (230)^{2} = 52900
P^{2} – 460P + 52900 = 42000 + 52900
(P – 230)^{2} = 10900
Step 4 Take the square root on both sides of the equation:
P – 230 = ±√10900 = ±104 (to nearest whole number)
Step 5 Subtract (230) from both sides (in other words, add 230):
P = 230 ± 104 = 126 or 334
What does that tell us? It says that the profit will be ZERO when the Price is $126 or $334
But we want to know the maximum profit, don't we?
It will be exactly half way inbetween! At $230
And here is the graph:
Profit = 200P^{2} + 92,000P  8,400,000
The optimum sale price is $230, and you can expect:
 Unit Sales = 70,000  200 x 230 = 24,000
 Sales in Dollars = $230 x 24,000 = $5,520,000
 Costs = 700,000 + $110 x 24,000 = $3,340,000
 Profit = $5,520,000  $3,340,000 = $2,180,000
  
A very profitable venture.
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EXAMPLE 3
Example: Small Steel Frame

Your company is going to make frames as part of a new product they are launching.
The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm^{2}
The inside of the frame has to be 11 cm by 6 cm
What should the width x of the metal be?

Area of steel before cutting:
Area = (11 + 2x) × (6 + 2x) cm^{2}
Area = 66 + 22x + 12x + 4x^{2}
Area = 4x^{2} + 34x + 66
Area of steel after cutting out the 11 × 6 middle:
Area = 4x^{2} + 34x + 66  66
Area = 4x^{2} + 34x
Let us solve this one graphically!
Here is the graph of 4x^{2} + 34x :
The required area of 28 is shown as a horizontal line.
The area equals 28 cm^{2} when:
x is approximately 9.3 or 0.8
The negative value of x make no sense, so the answer is:
x = 0.8 cm (approx.)
  
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EXAMPLE 4
Example: River Cruise
A 3 hour river cruise goes 15 km upstream and then back again. The river has a current of 2 km an hour. What is the boat's speed and how long was the upstream journey?

There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:
 Let x = the boat's speed in the water (km/h)
 Let v = the speed relative to the land (km/h)
Because the river flows downstream at 2 km/h:
 when going upstream, v = x2 (its speed is reduced by 2 km/h)
 when going downstream, v = x+2 (its speed is increased by 2 km/h)
We can turn those speeds into times using:
time = distance / speed
(if you travel 8 km at 4 km/h it would take 8/4 = 2 hours, right?)

And we know the total time is 3 hours:
total time = time upstream + time downstream = 3 hours
Put all that together:
total time = 15/(x2) + 15/(x+2) = 3 hours
Now we use our algebra skills to solve for "x".
First, get rid of the fractions by multiplying through by (x2)(x+2):
3(x2)(x+2) = 15(x+2) + 15(x2)
Expand everything:
3(x^{2}4) = 15x+30 + 15x30
Bring everything to the left and simplify:
3x^{2}  30x  12 = 0
Where a, b and c are from the
Quadratic Equation in "Standard Form": ax^{2} + bx + c = 0
Solve 3x^{2}  30x  12 = 0
Coefficients are:   a = 3, b = 30 and c = 12 
  
Quadratic Formula:   x = [ b ± √(b^{2}4ac) ] / 2a 
  
Put in a, b and c:   x = [ (30) ± √((30)^{2}4×3×(12)) ] / (2×3) 
  
Solve:   x = [ 30 ± √(900^{}+144) ] / 6 
  x = [ 30 ± √(1044) ] / 6 
  x = ( 30 ± 32.31 ) / 6 
  x = 0.39 or 10.39 
Answer: x = 0.39 or 10.39 (to 2 decimal places)
x =0.39 makes no sense for this real world problem, but x = 10.39 is just perfect!
Answer: Boat's Speed = 10.39 km/h (to 2 decimal places)
And hence the upstream journey = 15 / (10.392) = 1.79 hours = 1 hour 47min
And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min