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Chemistry LibreTexts

15.S: 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

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

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.