Why can't ax^2+bx+c=0 be rearranged into an equation to solve for a, b , c?

For a simple quadratic equation, let's say in standard for is simply


ax^2+bx+c=0. Why cannot this be rearranged to solve for a, b , or c


when you have 2 values such as "a" and "b", "a" and "c", or "b"


and "c".





Why doesn't the equation hold true when you are trying to rearrange


to solve for missing variables?!





x= -b+- sqrt b^2-4ac / 2a that is for solving x


Why doesn't it hold true if you do the following for other


variables:





To solve for b: b= -ax^2+c / x


To solve for a: bx+c / x^2


to solve for c: c= -ax^2-bx

Why can't ax^2+bx+c=0 be rearranged into an equation to solve for a, b , c?
for clarity of discussion, let's call "x" a variable, and "a", "b", "c" parameters. your question can be restated as:





given ax^2+bx+c=0 and any 2 parameters, plus the zeros, find the 3rd parameter.





let's examine the 3 cases.


but first recall that f(x)=ax^2+bx+c represents a parabola


(or for the degenerate cases, a line or a point). the parameters specify the characteristics of the curve (location, orientation, shape).





case 1: b,c,x known


ax^2+bx+c=0


we can subtract bx+c


ax^2=-bx-c


in the general case, x could be zero and therefore we cannot divide,








{standby, i'll try to finish this later -- got called away}
Reply:a, b, and c are NOT variables in a quadratic equation. x and y are the ONLY variables. a, b, and c represent CONSTANTS. Until all three constants are known, you DON'T have a quadratic equation.
Reply:You can solve for a,b,c if you have three points of the equation.





Since y = ax^2 + bx + c, you could plug the x and y values into the equation, and get three seperate equations where you could use a matrix to solve for the values of a, b, c.





I don't know if that really answers your question mainly b/c I don't fully understand your question. :/
Reply:they are constants not variables and you could replace them with variables and solve for them but then you would be solving an equation with more than one variable and it would be much more complicated, most likely with multiple solutions

wallflower