# Chapter 15.7: Essential Skills

- Page ID
- 28675

### 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 *b*^{2}/4*a*^{2}, which is equal to (*b*/2*a*)^{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, b^{2} ⁄ 4a^{2}, 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 4*a*^{2}. 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*,

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

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.

*x*^{2}+ 8*x*− 5 = 0- 2
*x*^{2}− 6*x*+ 3 = 0 - 3
*x*^{2}− 5*x*− 4 = 6 - 2
*x*(−*x*+ 2) + 1 = 0 - 3
*x*(2*x*+ 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