Question

Cite a real-life situation illustrating quadratic equations.

Answer 1

Answer:

Throwing a ball

Step-by-step explanation:

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)

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 position by about 5 m per second squared:   −5t2

(Note for the enthusiastic: the -5t2 is simplified from -(½)at2 with a=9.8 m/s2)    

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

h = 3 + 14t − 5t2

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

3 + 14t − 5t2 = 0

Which is a Quadratic Equation !

In "Standard Form" it looks like:

−5t2 + 14t + 3 = 0

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

5t2 − 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:

a×c = −15, and b = −14.

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)

Rewrite middle with −15 and 1: 5t2 − 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!

quadratic graph ball

Here is the graph of the Parabola h = −5t2 + 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.

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:

t = −b/2a = −(−14)/(2 × 5) = 14/10 = 1.4 seconds

Then find the height using that value (1.4)

h = −5t2 + 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.

Answer 2

Answer:

The polynomial equation whose highest degree is two is named a quadratic equation.

Step-by-step explanation:

To find the real-life situation illustrating quadratic equations.

Example: A company is going to make framings as part of a unique product they are undertaking. The frame will be cut out of a piece of steel, and to keep the weight down, the last area should be $28cm^{2}$. The interior of the framing has to be 11 cm by 6 cm.

Step 1

Area of steel before cutting $= (11 + 2x) * (6 + 2x) cm^2$

$= 66 + 22x + 12x + 4x^2$

$= 66+34x+4x^{2}$

Then we get,

Quadratic equation $= 4x^2 + 34x + 66$

Step 2

Area of steel after cutting out the 11 × 6 middle $= 4x^2 + 34x + 66 - 66$

$=4x^{2} +34x+0$

Then we get,

Quadratic Equation $= 4x^2 + 34x$

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