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

Question

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

Say if your given the zeroes and one variable, could the original
equation be rearranged into an new equation to solve for that
particular variable, sortof what I did above. Why is it that you
need the zeros. What is the minimum amount of information you could
be given to solve for missing variables and what situations are
their that you can use to solve for missing variables. ( i.e, your
given one zero and the vertex, solve for the other zero and solve
for the "a" value in the equation, or your given the zeros and the y-
value of the vertex, find the equation of the parabola). What is the
minimum set of points given to get an equation for a quadratic?

explain thouroughly please, with examples, thanks.

I am wondering same for the linear equation f(x)=mx+b

why can't it be rearranged to solve for x :
x= -b/m, or for b: b=-mx, or m= -b/x

Answer 1

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}

Answer 2

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. :/

Answer 3

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.

Answer 4

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