Nonlinear calibration curve. Following the procedure in Box 4-2, find how many micrograms $\left(\mu g\right)$of protein are contained in a sample with a corrected absorbance of 0.350in Figure 4-13.

Let us consider an unknown corrected absorbance of 0.35 lies beyond the linear region in Figure 4-13.

All the data points with the quadratic equation can be fitted$y=-1.17·{10}^{-4}{x}^2+0.01858x-0.0007.$

To find the quantity of protein, replace the corrected absorbance into the equation above:

$0.350=-1.17·{10}^{-4}{x}^2+0.01858x-0.0007.$

This equation can be rearranged as shown below.

$1.17·{10}^{-4}{x}^2-0.01858x+0.3507=0.$

A quadratic equation of the form:

$a{x}^2+bx+c=0$

The possible solutions are:

role="math" localid="1663330121693" $$\begin{gathered} x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ =\frac{0.01858\pm \sqrt{0.{01858}^2-4.1.17·{10}^{-4}0.3507}}{2·1.17·{10}^{-4}}. \end{gathered}$$

Solve further:

role="math" localid="1663330107186" $$\begin{gathered} x_1=\frac{0.01858\pm \sqrt{0.{01858}^2-4.1.17·{10}^{-4}0.3507}}{2·1.17·{10}^{-4}} \\ =137. \end{gathered}$$

role="math" localid="1663330101627" $$\begin{gathered} x_2=\frac{0.01858\pm \sqrt{0.{01858}^2-4.1.17·{10}^{-4}0.3507}}{2·1.17·{10}^{-4}} \\ =21.9. \end{gathered}$$

Figure tells us that the correct choice is $21.9\mu g$, not$137\mu g$.