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

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SUMMARY

The discussion centers on the necessity of using the quadratic formula to solve the equation Ax + Bx² + C = 0. Participants clarify that while alternative manipulations, such as x(A + Bx) + C = 0, can be explored, they do not yield valid solutions unless C = 0. The quadratic formula is essential as it provides two solutions for x, which cannot be derived from the proposed manipulations. Ultimately, the consensus is that the quadratic formula remains the most reliable method for solving quadratic equations.

PREREQUISITES
  • Understanding of quadratic equations and their standard form.
  • Familiarity with the quadratic formula: x = (-B ± √(B² - 4AC)) / (2A).
  • Basic algebraic manipulation skills, including factoring and solving for variables.
  • Knowledge of properties of equations, specifically the zero product property.
NEXT STEPS
  • Study the derivation and applications of the quadratic formula in various contexts.
  • Explore factoring techniques for quadratic equations and their limitations.
  • Learn about the discriminant in the quadratic formula and its implications for the nature of solutions.
  • Investigate alternative methods for solving quadratic equations, such as completing the square.
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Students studying algebra, educators teaching quadratic equations, and anyone seeking to deepen their understanding of solving polynomial equations.

yougene
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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

?
 
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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
 

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