From Red to Blue: A Scientific Investigation of Algebraic Transformations

[Miike012]

I would like to know how the book went from the term highlited in red to the term in blue.

 

[Mark44]

I would like to know how the book went from the term highlited in red to the term in blue.

They are using the quadratic formula. The expression in blue is what's inside the radical in this formula, the expression b² - 4ac.

Here a = 1, b = 2(1 - y), and c = 6y - 11

 

[Miike012]

The book is saying... if 2> y> 6 then there will be real roots ( I think ).
What does this have to do with the original equation (x^2 +2x - 11)/(2(x-3)) ?

 

[Mark44]

The book is saying... if 2> y> 6

This doesn't make any sense, because 2 is not greater than 6. I think what they are saying is the either y < 2 OR y > 6.

then there will be real roots ( I think ).
What does this have to do with the original equation (x^2 +2x - 11)/(2(x-3)) ?

It has to do with the quantity in the radical in the quadratic formula. This quantity is 4(1 - y)² - 4(6y - 11) = 4(1 - 2y + y²) - 24y + 44
= 4 - 8y + 4y² - 24y + 44 = 4y² - 32y + 48
= 4(y² - 8y + 12) = 4(y - 2)(y - 6).

In order for there to be two real roots of the quadratic equation, the quantity under the radical has to be > 0. This means that 4(y - 2)(y - 6) > 0, which means that either y < 2 or y > 6. Notice that if y = 2 or y = 6, there will be only a single real root of the quadratic equation.

 

[Miike012]

When I took the original equation and set it eqaul to 2 I got (x - 1)^2
=6 I got (x - 5)^2

Thus when y = 2 there is only one root of 1
y = 6 there is only one root of 5...

Is this what they are saying? And why would this matter? There must be something I am not seeing.

 

[PeterO]

When I took the original equation and set it eqaul to 2 I got (x - 1)^2
=6 I got (x - 5)^2

Thus when y = 2 there is only one root of 1
y = 6 there is only one root of 5...

Is this what they are saying? And why would this matter? There must be something I am not seeing.

The last line in the book said, to the effect; "so y cannot have values between 2 and 6 - all other values are allowed".

The value 2 is NOT between 2 and 6

Had they said y could not take on values from 2 to 6 it would be different

between 2 and 6 means 2 < y < 6 or (2,6)

from 2 to 6 means 2<= y <= 6 or [2,6]

the (2,6) and [2,6] specifications assumes you are familiar with that technique of answer.

 

[PeterO]

The book is saying... if 2> y> 6 then there will be real roots ( I think ).
What does this have to do with the original equation (x^2 +2x - 11)/(2(x-3)) ?

firstly 2 > y > 6 is not a great statement, as it implies 2 is greater than 6.
The book did not actually say that. Firstly they used only words, but the symbol version was

y > 6 gives two roots
y < 2 gives two roots

Implication was
y = 6 gives two equal roots - sometimes called just one root
y = 2 gives two equal roots - sometimes called just one root

leading to 2 < y < 6 means no real roots - or no real value if you like.

Since they named the original expression y, this means the original expression "can have all values except as lie between 2 and 6",

 

[Miike012]

Yes I understand that. I just wanted to see what would happen if y = 2 or 6.

But what is the importance? Why would someone need to know this?

 

[Mark44]

From post #4.

Notice that if y = 2 or y = 6, there will be only a single real root of the quadratic equation.

 

[Miike012]

From post #4.

And this is important why? How is this applicable? How will knowing this information help me?

 

[PeterO]

And this is important why? How is this applicable? How will knowing this information help me?

Have a read of the original question to re-focus on what you were trying to find.

 

[Ray Vickson]

And this is important why? How is this applicable? How will knowing this information help me?

It is not designed to help you. We are just saying that y = 2 is an achievable value, as is y = 6. All values < 2 or > 6 are also achievable. In fact, for a value of y > 6, say y = 7 there will be two values of x that give you that value of (x^2 +2x - 11)/(2(x-3)). For y = 2 or y = 6, only one single value of x will be obtained.

RGV

 

[NascentOxygen]

How will it help you? It probably won't; not today, anyway.

How does it help others? Well, if you are designing/building a mechanical or electrical or fluidic system, its dynamic behaviour can be characterised by a binomial equation like what you are examining. If that binomial equation has no roots (more exactly, no real roots) then that system won't bounce around when it gets a jolt. For example, when you flick ON an electrical switch, you generally wouldn't want the switch contacts to bounce, which could cause the circuit to switch ON then OFF then back ON then OFF until if finally stays ON. Bad bounce like that could be dangerous, or shorten the life of the switch or the load.

Other examples include applications where you DO accept some bounce or overshoot, because this means the response to a sudden jolt is faster or more comfortable. For example, in the suspension of your car, when a tyre hits a bump you want the car body to bounce up and down a bit (just once is ideal), to make the ride safer (less likely to break an axle), and more comfortable to the passengers. So in this case, you don't want the describing binomial to have real roots, because real roots point to behaviour that will be sluggish, slower to respond, and not so comfortable or as safe.

If the equation is of the parabolic trajectory of an arrow fired skywards, then with 2 roots there are two locations it could intercept a passing duck, with no roots the duck is safe.

 

[Miike012]

Ok. So this what ever it is, I don't know what to call it, seems like a waste of space in an alg book.
I guess right now, seeing that I will be starting calc 1, will have not benefit to me.

 

[NascentOxygen]

Ok. So this what ever it is, I don't know what to call it, seems like a waste of space in an alg book.
I guess right now, seeing that I will be starting calc 1, will have not benefit to me.

Come back in 6 months and tell us whether you're right.

 

[Miike012]

Come back in 6 months and tell us whether you're right.

Lol ok, Ill leave a note on my mirror to remind me.