Help needed, rearranging polynomial for inverse equation

AI Thread Summary
To rearrange the polynomial equation y = ax^2 + bx + c into the form x = ?, completing the square is a recommended method. The quadratic formula can also be applied after isolating y on one side of the equation. It’s important to note that a quadratic function typically does not have an inverse unless its domain is restricted. The discussion highlights using tools like the Wolfram Equation Solver for step-by-step solutions. Overall, both completing the square and the quadratic formula are valid approaches to find x in terms of y.
Charij
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Hi, I need to rearrange an equation:

y = ax^2 + bx + c

to the form of:

x = ?

I'm not entirely sure how to go about this and the examples I've found require the equation to be in a different form. Any tips or a point in the right direction would be great!

Thanks in advance
 
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Complete the square.
 
Great thanks!

That in mind I've got:

x = \frac{\sqrt{y - c - \frac{b^{2}}{4a}} - \frac{b}{2\sqrt{a}}}{\sqrt{a}}
 
Do you know the quadratic formula, which is the default solution of ax² + bx + c = 0? Or is that what you are trying to prove here? Because if not, you can pull y to the other side of the equals sign and apply the quadratic formula.
 
Complete the square or use the quadratic formula.

Most people learn how to solve quadratic equations before they learn about "inverse functions". Also, at some point you will have a "plus or minus". Unless your domain is restricted, a quadratic function will NOT have an inverse function.
 
I'm actually writing a program that works out a, b and c, but then needs to work out x given y. I probably used the wrong terminology to describe something along the way ^^

The answer I first wrote was generated by getting the equation in the form of:

y = (dx + e)^2 + f

and then working out d, e and f. The wolfram example is much nicer solution though, and more efficient computer wise :)

Thanks a lot for the help!
 
If you already have a square you can solve for p = (dx + e) first.
 
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