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Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

Geoffrey W. Cox, Ph.D., ?

Section A.3 Quadratic equations

We will be solving quadratic equations as we solve differential equations. If we want to solve a quadratic equation like \(ax^2 + bx + c = 0,\) there are several different methods we might use, including:
  1. factoring
  2. quadratic formula, \(\ds x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
  3. completing the square
Most students prefer the first two methods, which is fine. We will end up completing the square later in the semester, so if you want to review that method now, you’ll reap the benefits later!
Solve the following quadratic equations. Note: It’s OK if the solutions are complex or imaginary.
  1. \(\ds 4x^2 + 12x + 9 = 0 \qquad\)
    Solution.
    You might solve via factoring:
    \begin{align*} 4x^2 + 12x + 9 =\amp\ 0 \\ (2x + 3)(2x + 3) =\amp\ 0 \\ 2x+3 = 0 \amp \quad\text{or}\quad 2x+3 = 0 \\ x = -\frac{3}{2}, \amp \quad\text{or}\quad x = -\frac{3}{2} \\ \amp \\ \Rightarrow x = -\frac{3}{2} \text{ (double root)} \amp \end{align*}
    Alternately, you might use the quadratic formula:
    \begin{align*} x =\amp\ \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ =\amp\ \frac{-12 \pm \sqrt{12^2 - 4(4)(9)}}{2(4)} \\ =\amp\ \frac{-12 \pm \sqrt{144 - 144}}{8} \\ =\amp\ \frac{-12 \pm \sqrt{0}}{8} \qquad (\sqrt{0} \text{ implies a double root}) \\ =\amp\ -\frac{12}{8} \\ =\amp\ - \frac{3}{2} \end{align*}
    You could even complete the square:
    \begin{align*} 4x^2 + 12x + 9 =\amp\ 0 \\ 4x^2 + 12x =\amp\ -9 \\ x^2 + 3x =\amp\ -\frac{9}{4} \\ x^2 + 3x + \frac{9}{4} =\amp\ -\frac{9}{4} + \frac{9}{4}\\ \left(x + \frac{3}{2}\right)^2 =\amp\ 0 \\ x + \frac{3}{2} =\amp\ \pm\sqrt{0} \\ x =\amp\ -\frac{3}{2} \text{ (double root) } \end{align*}
    Answer.
    \begin{equation*} x = -\frac{3}{2} \mbox{ (double root)} \end{equation*}
  2. \(\ds 2x^2 - 9x - 35 = 0 \qquad\)
    Solution.
    You might solve via factoring:
    \begin{align*} 2x^2 - 9x - 35 =\amp\ 0 \\ (2x + 5)(x - 7) =\amp\ 0 \\ 2x+5 = 0 \amp \quad\text{or}\quad x-7 = 0 \\ x = -\frac{5}{2}, \amp \quad\text{or}\quad x = 7 \\ \amp \\ \Rightarrow x = -\frac{5}{2}, 7 \amp \end{align*}
    Alternately, you might use the quadratic formula:
    \begin{align*} x =\amp\ \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ =\amp\ \frac{-(-9) \pm \sqrt{(-9)^2 - 4(2)(-35)}}{2(2)} \\ =\amp\ \frac{9 \pm \sqrt{81 + 280}}{4} \\ =\amp\ \frac{9 \pm \sqrt{361}}{4} \\ =\amp\ \frac{9 \pm 19}{4} \\ =\amp\ \frac{9 + 19}{4}, \frac{9 - 19}{4} \\ =\amp\ \frac{28}{4}, \frac{-10}{4} \\ =\amp\ 7, -\frac{5}{2} \end{align*}
    You could even complete the square:
    \begin{align*} 2x^2 - 9x - 35 =\amp\ 0 \\ 2x^2 - 9x =\amp\ 35 \\ x^2 - \frac{9}{2}x=\amp\ \frac{35}{2} \\ x^2 - \frac{9}{2}x + \left( \frac{9}{4} \right)^2 =\amp\ \frac{35}{2} + \left(\frac{9}{4}\right)^2 \\ \left(x - \frac{9}{4}\right)^2 =\amp\ \frac{280}{16} + \frac{81}{16} \\ \left(x - \frac{9}{4}\right)^2 =\amp\ \frac{361}{16} \\ x - \frac{9}{4} =\amp\ \pm \sqrt{\frac{361}{16}} \\ x - \frac{9}{4} =\amp\ \pm \frac{19}{4} \\ x =\amp\ \frac{9}{4} \pm \frac{19}{4} \\ x =\amp\ \frac{9}{4} + \frac{19}{4}, \frac{9}{4} - \frac{19}{4} \\ x =\amp\ \frac{28}{4}, - \frac{10}{4} \\ x =\amp\ 7, - \frac{5}{2} \end{align*}
    Answer.
    \begin{equation*} x = -\frac{5}{2}, 7 \end{equation*}
  3. \(\ds x^2 - 4x + 13 = 0 \qquad\)
    Solution.
    This one doesn’t factor easily...
    You might use the quadratic formula:
    \begin{align*} x =\amp\ \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ =\amp\ \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(13)}}{2(1)} \\ =\amp\ \frac{4 \pm \sqrt{16 - 52}}{2} \\ =\amp\ \frac{4 \pm \sqrt{-36}}{2} \\ =\amp\ \frac{4 \pm 6i}{2} \\ =\amp\ 2 \pm 3i \\ =\amp\ 2 + 3i, 2 - 3i \end{align*}
    You could even complete the square:
    \begin{align*} x^2 - 4x + 13 =\amp\ 0 \\ x^2 - 4x =\amp\ -13 \\ x^2 - 4x + 4 =\amp\ -13 + 4 \\ (x-2)^2 =\amp\ -9 \\ x - 2 =\amp\ \pm \sqrt{-9} \\ x - 2 =\amp\ \pm 3i \\ x =\amp\ 2 \pm 3i \\ =\amp\ 2 + 3i, 2 - 3i \end{align*}
    Answer.
    \begin{equation*} x = 2 \pm 3i \end{equation*}
  4. Name at least two methods for solving quadratic equations.
    Answer.
    factoring, using the quadratic formula, completing the square
  5. How many solutions does a quadratic equation have?
    Answer.
    There are three possible outcomes when solving a quadratic equation:
    1. two distinct real roots
    2. one repeated real root (i.e., a double root)
    3. complex conjugate roots

Solving Quadratic Equations.

The solution to the quadratic equation
\begin{equation} a x^2 + b x + c = 0 \tag{A.1} \end{equation}
is given by the quadratic formula:
\begin{equation} x = \frac{-b \pm \sqrt{\color{blue} b^2 - 4ac}}{2a}\text{.}\tag{A.2} \end{equation}
Notes:
  1. The \(\pm\) gives two solutions, say \(x_1\) and \(x_2\text{.}\)
  2. \(x_1\) and \(x_2\) are also known as the roots of \(\ a x^2 + b x + c\text{.}\)
  3. The value, \({\color{blue} b^2 - 4ac} \ \text{,}\) under the root in is called the discriminant.
  4. Equation (A.1) can be written as \(\quad\ds (x - x_1)(x - x_2) = 0 \text{.}\)
  5. If \({\color{blue} b^2 - 4ac > 0}\text{,}\) then \(x_1\) and \(x_2\) are different real numbers.
  6. If \({\color{blue} b^2 - 4ac = 0}\text{,}\) then \(x_1\) and \(x_2\) are the same real number (repeated).
  7. If \({\color{blue} b^2 - 4ac < 0}\text{,}\) then \(x_1\) and \(x_2\) are complex and can be written as
    \begin{equation*} x_1 = \alpha + \beta i, \quad x_2 = \alpha - \beta i\text{.} \end{equation*}
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