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

$a_nx^n + ... + a_2x^2 + a_1x + a_0 \tag{15S.1}$

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:

$ax^2 + bx + c = 0\tag{15S.2}$

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 $$a$$:

$x^2+bax =−ca\tag{15S.3}$

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

Left side: $x^2+bax+b^24a^2=(x+b^2a)^2 \tag{15S.4}$

Having added a value to the left side, we must now add that same value, $$b^2 ⁄ 4a^2$$, to the right side:

$(x+b^2a)^2=−ca+b24a^2 \tag{15S.5}$

The common denominator on the right side is $$4a^2$$. Rearranging the right side, we obtain the following:

$(x+b^2a)^2=b^2−4ac4a^2 \tag{15S.6}$

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

$x+b^2a= \dfrac{\pm \sqrt{b^2−4ac}}{2a} \tag{15S.7}$

$x= \dfrac{−b \pm \sqrt{b^2−4ac}}{2a} \tag{15S.8}$

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

$x= \dfrac{−b + \sqrt{b^2−4ac}}{2a} \tag{15S.9}$
and
$x= \dfrac{−b - \sqrt{b^2−4ac}}{2a} \tag{15S.10}$

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. Exercise 15S.1 gives you practice using the quadratic formula.

Exercise 15S.1

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

1. $$x^2 + 8x − 5 = 0$$
2. $$2x^2 − 6x + 3 = 0$$
3. $$3x^2 − 5x − 4 = 6$$
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 $$x=−1−12−4(2)(−3)√2(2)=1.50$$

### Cubic Equations

The form of the cubic (i.e., 3rd degree) polynomial is:

$ax^3 + bx^2 + cx + d = 0$

which has three roots.

$x_1 = -\dfrac{b}{3a} - \dfrac{2^{1/3}(-b^2 + 3ac)}{3a(-2b^3 + 9abc - 27a^2d + \sqrt{4(-b^2 + 3ac)^3 + (-2b^3 + 9abc - 27a^2d)^2})^{1/3}} \\ + \dfrac{(-2b^3 + 9abc - 27a^2d + \sqrt{4(-b^2 + 3ac)^3 + (-2b^3 + 9abc - 27a^2d)^2})^{1/3}}{3 \cdot 2^{1/3}a}$

$x _2= -\dfrac{b}{3a} + \dfrac{(1 + i\sqrt{3})(-b^2 + 3ac)}{(3\cdot 2^{2/3}a(-2b^3 + 9abc - 27a^2d + \sqrt{4(-b^2 + 3ac)^3 + (-2b^3 + 9abc - 27a^2d)^2})^{1/3}} \\ - \dfrac{(1 - i\sqrt{3})(-2b^3 + 9abc - 27a^2d + \sqrt{4(-b^2 + 3ac)^3 + (-2b^3 + 9abc - 27a^2d)^2})^{1/3}}{6 \cdot 2^{1/3}a}$

$x_3 = -\dfrac{b}{3a} + \dfrac{(1 - i\sqrt{3})(-b^2 + 3ac)}{(3 \cdot 2^{2/3}a(-2b^3 + 9abc - 27a^2d + \sqrt{4(-b^2 + 3ac)^3 + (-2b^3 + 9abc - 27a^2d)^2})^{1/3}} \\- \dfrac{ (1 + i\sqrt{3}) \left(-2b^3 + 9abc - 27a^2d + \sqrt{4(-b^2 + 3ac)^3 + (-2b^3 + 9abc - 27a^2d)^2}\right)^{1/3}}{6 \cdot 2^{1/3}a}$

It is far easier to numerically solve cubic equations rather than to manipulate these unwieldy solutions.