Is there a maths genius on the forum?

Spuddy

Spuddy

Spuddy

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I'm having a nightmare with some math problems, if anyone is able to help I would very much appreciate it.

I've been tasked with plotting a graph with results of an experiment I conducted and then create a polynomial regression line. No problem.

I've created a graph and got my polynomial equation. The graph is as follows:
  • X axis = concentration
  • Y axis = absorbance

I also tested some samples of unknown concentration, but I have their Y value as I measured their absorbance level. Now I just need to work out what their concentration was.

How on earth do I find the X value given that I know the Y value? I'm clueless with maths like this.


Example of some samples and my equation:

Sample A = 3.0 absorbance
Sample B = 1.7275 absorbance

Equation:

Y = -0.7731x^2 + 3.0458x + 0.034


Obviously the Y value is changing with each sample so I'm confused how Y always equals the equation above, because Y is not constant.


Help pleeeeease.
 
You rearrange the equation to make x the subject. You already know Y for an unknown X, using your example above:

Y = -0.7731x^2 + 3.0458x + 0.034

Y = 3.0 (for A)

Therefore:

3.0 = -0.7731x^2 + 3.0458x + 0.034

Make X the subject (move it to where the 3.0 is and the 3.0 into the other side)

See if you can do it from there. If you need any more help let me know.
 
shredgodxp said:
You rearrange the equation to make x the subject. You already know Y for an unknown X, using your example above:

Y = -0.7731x^2 + 3.0458x + 0.034

Y = 3.0 (for A)

Therefore:

3.0 = -0.7731x^2 + 3.0458x + 0.034

Make X the subject (move it to where the 3.0 is and the 3.0 into the other side)

See if you can do it from there. If you need any more help let me know.
Click to expand...


Thank you, although I'm still unsure on how to do that re-arranging properly without watching a billion YouTube tutorials leading into the early AM.

I did try plugging in this equation into Wolfram Alpha but it gave me two plotted answers on the graph which just confused me even more.

3.0 = -0.7731x^2 + 3.0458x + 0.034


Another prod in the right direction would be great, please?
 
I like shredgodxp's hint so will continue in the same vein. :p

'Rearranging' is a bit of a red herring here as what you really want is the 'Quadratic formula'

https://en.wikipedia.org/wiki/Quadratic_formula

If you can make your equation look like the first one in that wiki page (ax^2 + bx + c = 0, tip: the 0 is the only difference!) then you just use the x = ... expression and bang, you have the answer. (Edit: or 'answers', hint hint, notice the plus/minus sybol, that means try both, plus or minus.)
 
joeyjojo said:
I like shredgodxp's hint so will continue in the same vein. :p

'Rearranging' is a bit of a red herring here as what you really want is the 'Quadratic formula'

https://en.wikipedia.org/wiki/Quadratic_formula

If you can make your equation look like the first one in that wiki page (ax^2 + bx + c = 0, tip: the 0 is the only difference!) then you just use the x = ... expression and bang, you have the answer.
Click to expand...

Ah yeah forgot bout that.

In another note, I'm guessing this is an analytical chemistry experiment. I'm used to working with linear graphs (working in the pharma industry means we need linear stuff) got stuck in that mindset.

But yeah follow the above hint and you'll get there!
 
Haha I sort of hate you guys but love you at the same time.


joeyjojo said:
I like shredgodxp's hint so will continue in the same vein. :p

'Rearranging' is a bit of a red herring here as what you really want is the 'Quadratic formula'

https://en.wikipedia.org/wiki/Quadratic_formula

If you can make your equation look like the first one in that wiki page (ax^2 + bx + c = 0, tip: the 0 is the only difference!) then you just use the x = ... expression and bang, you have the answer. (Edit: or 'answers', hint hint, notice the plus/minus sybol, that means try both, plus or minus.)
Click to expand...


Yeah that's what confused me before, I 'cheated' and used Wolfram Alpha earlier before posting here (after using an Excel formula) and it gave me two answers.

The two answers is what threw me off, how do I know which is the correct answer as my sample can only have one concentration to give me my absorbance value, right?

Also, it seems like my excel sheet did work but only gives one answer (the highest one of the two given by Wolfram Alpha).

Hope I'm not talking gibberish there?


shredgodxp said:
Ah yeah forgot bout that.

In another note, I'm guessing this is an analytical chemistry experiment. I'm used to working with linear graphs (working in the pharma industry means we need linear stuff) got stuck in that mindset.

But yeah follow the above hint and you'll get there!
Click to expand...

Yeah it was a protein quantification experiment using photo-spectrometry. Where do you work?

Thanks, I hope I get there :confused: haha. My brain is fried currently as I'm also revising for an exam! Running on empty.
 
Spuddy said:
The two answers is what threw me off, how do I know which is the correct answer as my sample can only have one concentration to give me my absorbance value, right?
Click to expand...

You get two answers because there are two 'sides' to a parabola. Draw a U shape and a horizontal line and they will cross at two points (or none if you draw it below, edit: or one if you get it precisely on the zenith).

Chances are one of them will be nonsense, because whatever experiment you're doing is only 'one side' of the parabola (the right side or the left side but not a U).

So the only way to know which is the right one is to look at how it fits in with the data you've measured.
 
Last edited:
A quadratic expression always has 2 answers unless you are looking at the peak/trough.

If there should only be 1 answer, you need to impose limits to narrow it down, or use common sense to deduct which is the red herring.

The experiment sounds like something in tertiary or higher education, but you need help on very early secondary maths education? What am I missing?
 
joeyjojo said:
You get two answers because there are two 'sides' to a parabola. Draw a U shape and a horizontal line and they will cross at two points (or none if you draw it below).

Chances are one of them will be nonsense, because whatever experiment you're doing is only 'one side' of the parabola (the right side or the left side but not a U).

So the only way to know which is the right one is to look at how it fits in with the data you've measured.
Click to expand...


Ok thank you, it seems as if the left hand side of that parabola is what fits the data best in the couple of samples I ran.

Will that be the case for all of them using this same data set then?

Thanks for your help btw, always appreciated.


D3K said:
The experiment sounds like something in tertiary or higher education, but you need help on very early secondary maths education? What am I missing?
Click to expand...

I'm an MSc Oncology student. This experiment was simply a 'practise run' for a method that I had not used for several years, but will be using more often in the coming months as part of a research project.

Being a BioMed student, I've not done a lot of maths since back in school as its not really required and as such I've forgot some of the more basic stuff. That's what you're missing.
 
joeyjojo said:
You get two answers because there are two 'sides' to a parabola. Draw a U shape and a horizontal line and they will cross at two points (or none if you draw it below).

Chances are one of them will be nonsense, because whatever experiment you're doing is only 'one side' of the parabola (the right side or the left side but not a U).

So the only way to know which is the right one is to look at how it fits in with the data you've measured.
Click to expand...

And that is why I prefer working with linear calibration curves. (Or at least applying a linear line of best fit and ensuring I have a suitable r2 value and good residual plots!)
 
This is the sort of stuff you learn in school, but 99% don't use it and therefore forget how to use it.

With age it's hard enough remembering if I locked the front door in the morning!
 
You may find it useful to get a hold of a program called Mathcad. It's basically a calculator that does algebra (and far more complex stuff as well).
 
shredgodxp said:
Ah yeah forgot bout that.

In another note, I'm guessing this is an analytical chemistry experiment. I'm used to working with linear graphs (working in the pharma industry means we need linear stuff) got stuck in that mindset.

But yeah follow the above hint and you'll get there!
Click to expand...

this is a linear model
 
Reminds me of when I had a maths gameboy. Didnt play games just drew cool graphs, I imagine thats relevant to this problem but I forgot how to use it now :o Surely a modern mobile will do stuff like this now
 
Rearranging you get:

Y = 1.96986 +/- 0.000129349 * sqrt((2.34551 * 10^8) - (7.731 * 10^7 * Y))

Yeah, you're not going to have much fun with this one, iterative? :)
 
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