Do you need to use the quadratic formula to solve for the unkown?

• yougene

Homework Statement

Lets say you have Ax + Bx^2 + C = 0

Usually I would use the Quadratic formula in this situation.

But what if you do this
x( A + Bx ) + C = 0
x( A + Bx ) = -C

Couldn't you then say
x = -C
and
A + Bx = -C
x = (-C - A) / B

?

yougene said:

Homework Statement

Lets say you have Ax + Bx^2 + C = 0

Usually I would use the Quadratic formula in this situation.

But what if you do this
x( A + Bx ) + C = 0
x( A + Bx ) = -C

Couldn't you then say
x = -C
and
A + Bx = -C
x = (-C - A) / B

?

Well you could say that, but it won't be correct. If you sub x=-C you will get

-C(A-BC) = -AC+BC2

and that is not the right side of the equation in red.

No. Factoring takes advantage of the property that if ab=0, then either a=0 or b =0. Furthermore, your answers don't solve the quadratic. Just plug them in.

And to extend what Random Variable said, if you had ab = 1, then all you know about a and b is that they are reciprocals of one another. They could be 1 and 1, -1 and -1, 2 and 1/2, $\sqrt{2}$ and $1/\sqrt{2}$, or any other of an infinite number of pairs of numbers that multiply to 1.

Thanks, this is all coming back now. :P

So then my equation would work only if C = 0.

@Mark44

So if my C = -1 then saying
Ax^2 + Bx = 1
would be a valid way to solve the equation?

Or are you saying
Ax^2 + Bx + C = 1
would be valid to solve for?

yougene said:
Couldn't you then say
x = -C
and
A + Bx = -C

No. Why do you think that? All that is telling you is that x and (A+Bx) are always -C when multiplied together if x is a solution to the quadratic formula. That doesn't tell you anything about what x actually is. Remember, the quadratic formula has 2 solutions so you have to get 2 numbers out of it for x.

yougene said:
So then my equation would work only if C = 0.

I believe that would be correct. If you had
x (A+Bx) = 0

then either (x) or (A+Bx) must be zero, or both. The two solutions would be
x =0 and
x=-A/B