Need help with quadratic equation

Factual Questions
1

I’ve been busy building an enzymatic assay to measure a protein. The data points of the standards used to create a calibration curve, when graphed, fit perfectly on a second order polynomial curve (i.e. aX^2+bX+c). I remember the formula for solving for X when Y=0, however, we’re in the real world, and I need to find X when, say, 1.225=aX^2+bX+c. From Excel, I know a, b, and c, and from my experimental data, I know Y. What I need to know, and can’t find, is how to solve for X when Y has an explicit value not equal to 0. Any help from those more mathematically inclined than I would be greatly appreciated.

Vlad/Igor

2

Set the equation to aX^2+bX+(c-Y)=0 and use the QE

3

I tried that and got the following, but I don’t know if it correct:

X = -b+/- sqrt(4a(Y-c)+b^2)/2a

Vlad/Igor

4

That’s right. You can always substitute the values back in the original equation to check.

5

Your answer is right, assuming that you implicitly have parentheses as follows:

X = (-b+/- sqrt(4a(Y-c)+b^2))/(2a)

That is, the 2a needs to divide the entire sum -b+/- sqrt(4a(Y-c)+b^2), not just the square root.

6

Shouldn’t it be x=(-b+/-sqrt(b[sup]2[/sup]**-**4ac))/(2a) where c=c-Y?

7

c = c - Y implies Y = 0.

8

I think he meant a new c, c[sub]2[/sub], where c[sub]2[/sub] = c[sub]1[/sub] - Y

9

Yes, but that’s equivalent because replacing c with Y-c instead of c-Y cancels the minus sign.