polynomial equation

[PAUL GRAZIDE]

I am setting up a calibration curve, where x= concentration and y=
absorbance. The curve is polynomial.

after plotting x and y values using a scattergraph and applying a trend
line, i got the following regression line:
y= -0.0881x^2+2.8089x+0.0505.

If the absorbance reading (y) = 0.212, how do I calculate the value of x?

Thanks in advance

 

[davesexcel]

0.057672821
y= -0.0881x^2+2.8089x+0.0505.

If the absorbance reading (y) = 0.212, how do I calculate the value of
x?
I am just thinking
A1=B2
B2=-.0881*C1^2+2.8089*C1+.0505
then I went to Tools, Goal seek
set the value of A1 to .212 by changing Cell C1

 

[Guest]

Just solve the quadratic equation:

0= -0.0881x^2+2.8089x-0.1615

x=31.82549

see:

http://mathworld.wolfram.com/QuadraticEquation.html

 

[davesexcel]

Just solve the quadratic equation:

0= -0.0881x^2+2.8089x-0.1615

x=31.82549

see:

http://mathworld.wolfram.com/QuadraticEquation.html
--
Gary''s Student


"

[/QUOTE]
Well I got that wrong Hmmnn... I wonder why the goal seek didn't work?

 

[Guest]

You are correct davesexcel:

This quadratic has two separate solutions:
one around .0576 and the other around 31.82548752

=(-2.8089+(2.8089^2-4*(-0.0881)*(-0.1615))^0.5)/(2*(-0.0881))
=(-2.8089- (2.8089^2-4*(-0.0881)*(-0.1615))^0.5)/(2*(-0.0881))

 

[Bernard Liengme]

Do NOT copy the polynomial coefficients from the trendline but use LINEST -
see
http://www.stfx.ca/people/bliengme/ExcelTips/Polynomial.htm
When you look at the quadratic solution, remember only positive x is
meaningful in this situation
best wishes from a chemist

 

[Guest]

Or if you do use the chart equation, format it to display scientific notation
with 14 decimal places to avoid rounding issues.

More fundamentally, are you sure that a 4- or 5-parameter logistic would not
be a better model for your data than a polynomial?

Jerry