15.7: Essential Skills 7

Last updated
Save as PDF

Page ID: 349752

Anonymous
LibreTexts

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

Learning Objectives

The quadratic formula

Previous Essential Skills sections introduced many of the mathematical operations you need to solve chemical problems. We now introduce the quadratic formula, a mathematical relationship involving sums of powers in a single variable that you will need to apply to solve some of the problems in this chapter.

The Quadratic Formula

Mathematical expressions that involve a sum of powers in one or more variables (e.g., x) multiplied by coefficients (such as a) are called polynomials. Polynomials of a single variable have the general form

\[a_nx^n + ... + a_2x^2 + a_1x + a_0 \tag{15.7.1} \]

The highest power to which the variable in a polynomial is raised is called its order. Thus the polynomial shown here is of the nth order. For example, if n were 3, the polynomial would be third order.

A quadratic equation is a second-order polynomial equation in a single variable x:

\[ax^2 + bx + c = 0 \tag{15.7.2} \]

According to the fundamental theorem of algebra, a second-order polynomial equation has two solutions—called roots—that can be found using a method called completing the square. In this method, we solve for x by first adding −c to both sides of the quadratic equation and then divide both sides by a:

\[x^2+\dfrac{bx}{a} =−\dfrac{c}{a} \tag{15.7.3} \]

We can convert the left side of this equation to a perfect square by adding b²/4a², which is equal to (b/2a)²:

Left side: \[x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}=(x+\dfrac{b}{2a})^2 \tag{15.7.4} \]

Having added a value to the left side, we must now add that same value, b² ⁄ 4a², to the right side:

\[(x+\dfrac{b}{2a})^2=−\dfrac{c}{a}+\dfrac{b^2}{4a^2} \tag{15.7.5} \]

The common denominator on the right side is 4a². Rearranging the right side, we obtain the following:

\[(x+\dfrac{b}{2a})^2=\dfrac{b^2−4ac}{4a^2} \tag{15.7.6} \]

Taking the square root of both sides and solving for x,

\[x+\dfrac{b}{2a}= \dfrac{\pm \sqrt{b^2−4ac}}{2a} \tag{15.7.7} \]

\[x= \dfrac{−b \pm \sqrt{b^2−4ac}}{2a} \tag{15.7.8} \]

This equation, known as the quadratic formula, has two roots:

\[x= \dfrac{−b + \sqrt{b^2−4ac}}{2a} \tag{15.7.9} \]

and

\[x= \dfrac{−b - \sqrt{b^2−4ac}}{2a} \tag{15.7.10} \]

Thus we can obtain the solutions to a quadratic equation by substituting the values of the coefficients (a, b, c) into the quadratic formula.

When you apply the quadratic formula to obtain solutions to a quadratic equation, it is important to remember that one of the two solutions may not make sense or neither may make sense. There may be times, for example, when a negative solution is not reasonable or when both solutions require that a square root be taken of a negative number. In such cases, we simply discard any solution that is unreasonable and only report a solution that is reasonable. Skill Builder ES1 gives you practice using the quadratic formula.

Skill Builder ES1

Use the quadratic formula to solve for x in each equation. Report your answers to three significant figures.

x² + 8x − 5 = 0
2x² − 6x + 3 = 0
3x² − 5x − 4 = 6
2x(−x + 2) + 1 = 0
3x(2x + 1) − 4 = 5

Solution:

\(9x=−8+82−4(1)(−5)√2(1)=0.583\) and \(x=−8−82−4(1)(−5)√2(1)=−8.58\)
\(x=−(−6)+(−62)−4(2)(3)√2(2)=2.37\) and \(x=−(−6)−(−62)−4(2)(3)√2(2)=0.634\)
\(x=−(−5)+(−52)−4(3)(−10)√2(3)=2.84\) and \(x=−(−5)−(−52)−4(3)(−10)√2(3)=−1.17\)
\(x=−4+42−4(−2)(1)√2(−2)=−0.225\) and \(x=−4−42−4(−2)(1)√2((−2))=2.22\)
\(x=−1+12−4(2)(−3)√2(2)=1.00\) and \(x=−1−12−4(2)(−3)√2(2)=1.50\)

Contributors

Anonymous

Modified by Joshua B. Halpern