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## 15.7 Essential Skills

### Topic

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

anxn + ··· + a2x2 + a1x + a0

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:

ax2 + bx + c = 0

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:

$x 2 + b a x = − c a$

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

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

$( x + b 2 a ) 2 = − c a + b 2 4 a 2$

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

$( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2$

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

$x + b 2 a = ± b 2 − 4 a c 2 a$ $x = − b ± b 2 − 4 a c 2 a$

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.

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. $x=−8+82−4(1)(−5)2(1)=0.583$ 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 $x=−1−12−4(2)(−3)2(2)=1.50$