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Help needed, rearranging polynomial for inverse equation

 
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Mar3-13, 02:06 PM   #1
 

Help needed, rearranging polynomial for inverse equation


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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Mar3-13, 02:15 PM   #2
 
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Complete the square.
 
Mar3-13, 02:54 PM   #3
 
Great thanks!

That in mind I've got:

x = [itex]\frac{\sqrt{y - c - \frac{b^{2}}{4a}} - \frac{b}{2\sqrt{a}}}{\sqrt{a}}[/itex]
 
Mar4-13, 09:53 AM   #4
 

Help needed, rearranging polynomial for inverse equation


Check out the Wolfram Equation Solver:

http://www.wolframalpha.com/examples...onSolving.html

You would want to "solve an equation with parameters". Their answer looks a bit different, so you can look at their step by step breakdown and see whether your answer is equivalent.
 
Mar4-13, 10:38 AM   #5
 
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Do you know the quadratic formula, which is the default solution of ax˛ + 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.
 
Mar4-13, 10:39 AM   #6
 
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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.
 
Mar4-13, 10:43 AM   #7
 
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!
 
Mar5-13, 03:18 AM   #8
 
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If you already have a square you can solve for p = (dx + e) first.
 
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