SOLUTION: When working out the rule for a quadratic equation (ax^2+bx+c)from a tables of different values how do you find B? I know that a=half second difference pattern and c=y when x=0. I

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: When working out the rule for a quadratic equation (ax^2+bx+c)from a tables of different values how do you find B? I know that a=half second difference pattern and c=y when x=0. I       Log On


   

Question 66047: When working out the rule for a quadratic equation (ax^2+bx+c)from a tables of different values how do you find B? I know that a=half second difference pattern and c=y when x=0. I covered it ages ago and its just gone straight out of my head lol.. Thanks Jocelyn
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
This is the quadratic equation. However i think you are asking how to find the middle term of the equation ax^2+bx+c=0
It generally is best to set a=1 by dividing all terms by a. Then set up the following table assuming we have the equation (x^2-10x+25).
Knowing that x+y=b & x*y=c we need to find:
The factors of (c) 25. They are (+1)(+25),(-1)(-25) & (+5)(+5),(-5)(-5)
Now find 2 of these factors that add up to (b) (-10). That would be
(-5)+(-5)=-10
So the factors of x^2-10x+25 are (x-5)(x-5).[-5-5=-10 & (-5)(-5)=25]
Another example is x^2-3x-10.
The factors of (c) -10 are (-1)(10),(1-10),(-2)(5),(2)(-5).
Now find 2 of these factors that add up to (b) (-3). That would be (2)+(-5)=-3.
So the factors here would be (x+2)(x-5). [2-5=-3 & (2)(-5)=-10]
Hope this helps.