# Chapter 15.7: Essential Skills

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 Howard University General Chemistry: An Atoms First Approach Unit 1: Atomic Theory      Unit 2: Molecular Structure        Unit 3: Stoichiometry          Unit 4: Thermochem & Gases Unit 5:  States of Matter         Unit 6: Kinetics & Equilibria        Unit 7: Electro & Thermo Chemistry       Unit 8: Materials

### Learning Objectives

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.

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.

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

Solution:

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

### Contributors

• Anonymous

Modified by Joshua B. Halpern