Thursday, July 9, 2009

Can the standard quadratic equation (Ax^2+bx+c=0) be rearranged to solve for other variables such as A,b,C?

Standard Equation of a parabola is:


ax^2+bx+c=0,


to rearrange to find for zeros its then turned into x=-b+- (sqrt b^2-4ac)/2a, but say If your given points on a graph say the y-interecept (c) and the a value of the equation, Could you solve for b by rearranging the equation? Or maybe if your given the zeroes and the y intercept, Could you solve for the rest of the missing variables such as "A" and "B" by rearranging the equation? Maybe if your also given the vertex and maybe one point, could you solve for missing variables? Can someone explain to me If the equation can be rearranged to solve for variables besides the zeroes (x)?, If so would it work out and how?





My Random Ideas:





ax^2+bx+c=0


-c=ax^2+bx





-ax^2=bx+c





-a=bx/x^2+c/x^2





-bx=ax^2+c

Can the standard quadratic equation (Ax^2+bx+c=0) be rearranged to solve for other variables such as A,b,C?
You certainly can. However - you should remember that the standard equation of a parabola is y=ax^2 +bx +c (not 0=...).





Plugging y=0 into this equation will just give the one or two points on the parabola that the graph touches the x-axis if there is a real solution - or tell us that the graph lies wholly above the x-axis if there are no real solutions.





But you can also solve for the set of a, b and c by plugging in value of x and y (need at least three sets of points if all of a, b and c are missing). Otherwise you need one value of points for each of a, b and c that is unknown.





So if you have three points on the x-y plane that fit the curve and you plug them in you will get a set of equations in terms of a, b, and c





For example say you have the set of points (0, 6), (1, 2) and (2,0)





(0, 6) =%26gt; 6 = a.0 + b.0 + c (1)


(1, 2) =%26gt; 2 = a.1 + b.1 + c (2)


(2, 0) =%26gt; 0 = a.4 + b.2 + c (3)





(1) =%26gt; c=6


(2) =%26gt; 2= a+b+c =%26gt; 2=a+b+6 =%26gt; a+b=-4


(3) =%26gt; 0=4a+2b+c =%26gt; 0=4a+2b+6 =%26gt; 2a+b=-3


Which gives a=1 and b=-5





Curve is y=x^2 -5x +6





They may not solve as nicely as this (I picked a nice values - always good to have the x-intercept since it automatically gives you the value of c) - but as long as the three points are on the parabola you will always be able to solve for a, b and c.





Your way would work (and is equivalent to saying we know 2 of the points are (x1, 0) and (x2, 0)) as long as you also knew at least one of a, b and c. Of course you will have a massive problem with your method if there are no real points where y=0.
Reply:Absolutely yes. If you know the other variables except for one other one then you can solve for it.
Reply:2 + 2 = 22





your welcome





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