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Introduzione interattiva alla formula quadratica

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Introduzione interattiva alla formula quadratica

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Utilizzare la LIM per proiettare una presentazione che introduce le equazioni di secondo grado e la formula risolutiva: ax^2 + bx + c = 0. Discutere ogni componente e il suo ruolo.

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Analisi grafica delle soluzioni delle equazioni

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Analisi grafica delle soluzioni delle equazioni

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Usare il software sul computer della classe per mostrare graficamente come i coefficienti a, b, e c influenzano le radici dell'equazione. Gli studenti possono manipolare i valori per vedere un risultato immediato.

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Esercitazione guidata: Risoluzione di equazioni quadrate

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Esercitazione guidata: Risoluzione di equazioni quadrate

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Fornire agli studenti una serie di equazioni di secondo grado da risolvere. Guidare il processo iniziale e quindi lasciare che gli studenti risolvano in piccoli gruppi.

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Gioco di squadra: Equazione quadratica a staffetta

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Gioco di squadra: Equazione quadratica a staffetta

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Dividere la classe in squadre. Ogni squadra deve risolvere un'equazione, passando una 'staffetta' al membro successivo che continua a risolvere l'equazione corretta successiva.

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Creare un puzzle dove pezzi separati contengono parti di equazioni di secondo grado da combinare correttamente. Gli studenti lavorano in coppia per rimettere insieme il puzzle.

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Preparare un quiz interattivo utilizzando software come Kahoot o Quizlet per ricapitolare le conoscenze sulla risoluzione delle equazioni di secondo grado.

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Laboratorio: Creazione di problemi reali

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Laboratorio: Creazione di problemi reali

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Ogni studente crea un problema della vita reale che può essere risolto con un'equazione quadratica, presentandolo poi alla classe per la soluzione collettiva.

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A Quadratic Equation looks like this:

Quadratic equations pop up in many real world situations!

Here we have collected some examples for you, and solve each using different methods:

Each example follows three general stages:

Take the real world description and make some equations

Solve!

Use your common sense to interpret the results

Balls, Arrows, Missiles and Stones

When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster ...

... and a Quadratic Equation tells you its position at all times!

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:
(Note: t is time in seconds)

Add them up and the height h at any time t is:

And the ball will hit the ground when the height is zero:

Which is a Quadratic Equation !

In "Standard Form" it looks like:

It looks even better when we multiply all terms by −1:

5t ² − 14t − 3 = 0

Let us solve it ...

There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give a×c, and add to give b" method in Factoring Quadratics:

The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15

By trying a few combinations we find that −15 and 1 work (−15×1 = −15, and −15+1 = −14)

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

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.

Also notice that the ball goes nearly 13 meters high.

Note: 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:

Then find the height using that value (1.4)

So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

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 set-up costs, advertising, etc

$110 to make each bike

Based on similar bikes, you can expect sales to follow this "Demand Curve":

Where "P" is the price.

For example, if you set the price:

at $0, you just give away 70,000 bikes

at $350, you won't 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 − 200P²

Costs = 700,000 + 110 x (70,000 − 200P) = 700,000 + 7,700,000 − 22,000P = 8,400,000 − 22,000P

Profit = Sales-Costs = 70,000P − 200P² − (8,400,000 − 22,000P) = −200P² + 92,000P − 8,400,000

Profit = −200P ² + 92,000P − 8,400,000

Yes, a Quadratic Equation. Let us solve this one by Completing the Square.

Solve: −200P² + 92,000P − 8,400,000 = 0

Step 1 Divide all terms by -200

Step 2 Move the number term to the right side of the equation:

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:

Step 4 Take the square root on both sides of the equation:

Step 5 Subtract (-230) from both sides (in other words, add 230):

What does that tell us? It says that the profit is ZERO when the Price is $126 or $334

But we want to know the maximum profit, don't we?

It is exactly half way in-between! At $230

And here is the graph:

Profit = −200P ² + 92,000P − 8,400,000

The best 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.

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 cm²

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 of steel after cutting out the 11 × 6 middle:

Area = 4x ² + 34x + 66 − 66

Let us solve this one graphically!

Here is the graph of :

The desired area of 28 is shown as a horizontal line.

The area equals 28 cm ² when:
x is about −9.3 or 0.8
The negative value of x make no sense, so the answer is:

x = 0.8 cm (approx.)

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 = x−2 (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

(to travel 8 km at 4 km/h takes 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/(x−2) + 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 (x-2)(x+2):

3(x-2)(x+2) = 15(x+2) + 15(x-2)

Expand everything:

3(x ² −4) = 15x+30 + 15x−30

Bring everything to the left and simplify:

3x ² − 30x − 12 = 0

It is a Quadratic Equation! Let us solve it using the Quadratic Formula:

Solve 3x² - 30x - 12 = 0

(to 2 decimal places)

x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect!

Answer: Boat's Speed = 10.39 km/h (to 2 decimal places)

And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min

And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min

Example: Resistors In Parallel

Two resistors are in parallel, like in this diagram:

The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other.

What are the values of the two resistors?

The formula to work out total resistance "R T" is:

To get rid of the fractions we can multiply all terms by 2R ₁(R ₁ + 3) and then simplify:

Yes! A Quadratic Equation !

Let us solve it using our Quadratic Equation Solver.

Enter 1, −1 and −6

And you should get the answers −2 and 3

The two resistors are 3 ohms and 6 ohms.

Others

Quadratic Equations are useful in many other areas:

For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation.

Quadratic equations are also needed when studying lenses and curved mirrors.

And many questions involving time, distance and speed need quadratic equations.

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Real World Examples of Quadratic Equations

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A Quadratic Equation looks like this: Quadratic equations pop up in many real world situations! Here we have collected some examples for you, and solve each using different methods: Each example follows three general stages: Take the real world description and make some equations Solve!

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Laboratorio informatico: Programmazione delle equazioni di secondo grado

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Laboratorio informatico: Programmazione delle equazioni di secondo grado

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Insegnare agli studenti a programmare una semplice applicazione che risolve equazioni quadrate. Utilizzare un linguaggio di programmazione base, come Python, con la supervisione dell'insegnante.

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Discussione: Il significato delle soluzioni delle equazioni

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Discussione: Il significato delle soluzioni delle equazioni

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Indirizzare una discussione sui diversi tipi di soluzioni delle equazioni quadrate (reali e immaginarie) e il loro significato nel contesto matematico ed applicato.

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Video didattico: Equazioni quadrate nel mondo reale

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Video didattico: Equazioni quadrate nel mondo reale

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Guardare un video che spiega come le equazioni di secondo grado vengono usate nella vita reale, ad esempio nell'ingegneria e nell'economia.