Linear and Quadratic equations

[autodidude]

This is something that's been bothering me for a while...

If quadratics are any polynomials that have a degree of 2, then would that make the following equations quadratics (I don't know what all of them are for by the way, they're just equations I'm pulling from my math book which doesn't say)

I = k/d^2

v = ω√(r^2 - y^2)

A=l^2

In the book, they're under linear functions...right now, I'm thinking that they are quadratics but only quadratics that are like x^2 + x must be solved using the various methods for solving quadratic equations, whereas if it's like x^2 + y, you can just rearrange and take the square root




I'm wrong, right? haha

 

[symbolipoint]

I = k/d^2

v = ω√(r^2 - y^2)

A=l^2

The first two can be converted to quadratic equations using multiplication properties of equality. The last one is quadratic. These assume that your variables are d, r, y, and the the last equation, l.

 

[HallsofIvy]

This is something that's been bothering me for a while...

If quadratics are any polynomials that have a degree of 2, then would that make the following equations quadratics (I don't know what all of them are for by the way, they're just equations I'm pulling from my math book which doesn't say)

I = k/d^2

If you mean I as a function of d, this is not a polynomial because the independent variable is in the denominator. This is a "rational" function.

v = ω√(r^2 - y^2)

If you mean v as a function of r or y, this is not a polynomial because of the square root.

A=l^2

Yes, A is a quadratic function of l

In the book, they're under linear functions...right now, I'm thinking that they are quadratics but only quadratics that are like x^2 + x must be solved using the various methods for solving quadratic equations, whereas if it's like x^2 + y, you can just rearrange and take the square root




I'm wrong, right? haha

If you mean I as a function of k and v as a function of ω then the first two are linear.

 

[autodidude]

What do you mean 'a function of'?

 

[jhae2.718]

What do you mean 'a function of'?

E.g., if we write y=ax², we say that y is a function of x. (Here a is a constant, and y is not a function of a.) Function notation: f(x), where f is the function and x is the variable it operates on. So, we could say y = f(x) = ax²

A function maps a set on inputs to a set of outputs. So, if the input to your function with the root is ω, then we would say it is a function of ω...

 

[autodidude]

^ I think I understand you...

This is high school level math right (specifically, Yr 11)? Where can I read more on this stuff (ie. more info on quadratics, what and what isn't a polynomial) Because the book I'm using (sure sure, blame the book :p). Most sites I've come across give pretty don't offer the information I'm getting here.

But I never completed Yr 10 math so that may explain why I'm missing a lot of key concepts

 

[jhae2.718]

That's algebra II? And yr 10 is...geometry?

 

[autodidude]

Yeah...I remember doing some basic Algebra and Trigonometry in Yr 10 but I never got up to quadratics

 

[autodidude]

Anybody else want to chip in?

 

[HallsofIvy]

Chip in what? Everything has been pretty much said. You started by saying "quadratics are any polynomials that have a degree of 2" and only the last, A= l^2, is a polynomial.

 

[autodidude]

I'm thinking that they are quadratics but only quadratics that are like x^2 + x must be solved using the various methods for solving quadratic equations, whereas if it's like x^2 + y, you can just rearrange and take the square root

:)

I'm quite slow

 

[Mark44]

I'm thinking that they are quadratics but only quadratics that are like x^2 + x must be solved using the various methods for solving quadratic equations, whereas if it's like x^2 + y, you can just rearrange and take the square root

A quadratic equation such as x² + x = 0[/color] can be solved by factoring, completing the square, or using the Quadratic Formula.

x² + x is an expression, not an equation, so it can't be solved. The best you can do is to write it in some other form that is identically equal to the original expression (e.g., x² + x = x(x + 1)). We haven't solved anything here.

x² + y is also just an expression, so there's really not much you can do with it. In particular, you can't rearrange it so as to solve it for y.

If you had an equation, x² + y = 0, then you could rearrange to an equivalent equation, y = - x². There wouldn't be much point of taking the square root of both sides, though.

 

[HallsofIvy]

I'm thinking that they are quadratics but only quadratics that are like x^2 + x must be solved using the various methods for solving quadratic equations, whereas if it's like x^2 + y, you can just rearrange and take the square root

:)

I'm quite slow

No, they are not quadratics. Again, as you said, a quadratic is a polynomial of degree 2. I = k/d^2 and v = ω√(r^2 - y^2) are NOT polynomials. A polynomial must involve only positive integer powers of the variables. I= k/d^2= kd^{-2} involves a negative power and v= \omega\sqrt{r^2- y^2}= \omega(r^2-y^2)^{1/2} involves a fractional power.

 

[autodidude]

A quadratic equation such as x² + x = 0[/color] can be solved by factoring, completing the square, or using the Quadratic Formula.

x² + x is an expression, not an equation, so it can't be solved. The best you can do is to write it in some other form that is identically equal to the original expression (e.g., x² + x = x(x + 1)). We haven't solved anything here.

x² + y is also just an expression, so there's really not much you can do with it. In particular, you can't rearrange it so as to solve it for y.

If you had an equation, x² + y = 0, then you could rearrange to an equivalent equation, y = - x². There wouldn't be much point of taking the square root of both sides, though.

My mistake, when I said x^2 + y and x^2 + x, I meant x^2 + y = z and x^2 + x = y, solve for x