the form of roots

[Vastor]

[math] ax^2 + bx + c = 0 [/math] is general form

 

by using factorisation, we can turn it into this form

 

let's say...

 

[math] (x - r_1)(x + r_2) = 0 [/math]

 

then we get the roots as follow

 

[math] x - r_1 = 0 [/math]

 

[math] x = r_1 [/math]

 

&

 

[math] x + r_2 = 0 [/math]

 

[math] x = - r_2 [/math]

 

compare to forming an equation out of roots(where roots are given) it's different story, where it's giving me some headache to think about...

 

usually, teacher will give formula

 

[math] x^2 - (S.O.R) + (P.O.R) [/math]

 

S.O.R = sum of root

P.O.R = product of root

 

but, I personally don't like remembering and plugin stuff in math, because that's not the way to learn it...

 

so, I use this equation instead...

 

[math] (x - r_1)(x + r_2) = 0 [/math] (an equation that i got from factorisation)

 

and this is the part where my math got stuck... let's say it's given roots x = 3 and x = -2

 

so, i move on and turn

 

x = 3

x - 3 = 0

 

but what does 0's value refer to?(i know it's y / f(x) but how does to proof that mathematically?)

 

so, trying to not break any math law, i write on my answer paper this instead

 

let y = 0 (because it's roots, so, its plot should be on the x intercept where y = 0)

x = 3 + y

x - 3 = y

x - 3 = 0

 

and, x + 2 = 0

 

another question is, what calculation u can proof that "(x - 3)(x + 2) = 0" can be form from "x - 3 = 0" and "x + 2 = 0" ?

 

i found this is so blurry, i know how to solve this all, but just don't get what math were used in this calculation, hope anyone can help...

[ajb]

The quadratic equation is given by

 

[math] f(x) = a x^{2} + b x + c[/math].

 

When you talk about "solutions" or "roots" (or whatever you call it) one is referring to setting [math]f(x)=0[/math] and then solving for [math]x[/math]. If not zero, them lets say you want [math]f(x)=d[/math] for some fixed number, then you are simply looking for solutions of

 

[math]0 =a x^{2} + b x + (c-d)[/math],

 

so solving for the quadratic equations for its' zeros is in reality quite general.

 

We need to be a little careful with what field we are working over. Lets say we are over the complex numbers as this field is algebraically closed. If you don't know about complex numbers, then just think of real numbers and we restrict our attentions to quadratic equations that have real roots. Then one can always write the quadratic equation as

 

[math]f(x) = (x-A)(x-B)[/math],

 

for two complex numbers. Just expand it out and you can match terms with the initial quadratic equation. (Do this, I think it will help you)

 

The only complex (or real) number [math]C[/math] say, such that [math]yC = 0[/math] for all complex (or real) numbers [math]y[/math] is [math]0[/math]. So in the above setting [math]x = A,B[/math] are the only numbers that produce [math]f(x)=0[/math].

 

I hope I have helped you in some way.

Edited September 11, 2011 by ajb