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5.3.3: Quadratic Formula and Complex Sums

  • Page ID
    4161
  • Solve quadratic equations with complex roots, and add and subtract complex numbers.

    Complex Roots of Quadratic Functions

    The quadratic function \(y=x^2−2x+3\) (shown below) does not intersect the x-axis and therefore has no real roots. What are the complex roots of the function?

    f-d_a9362114ed79659c0b67a36746efb0d37ee4ffe4e9b4817472d1c013+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{1}\)

    Complex Roots of Quadratic Functions

    Recall that the imaginary number, \(i\), is a number whose square is –1:

    \(\textcolor{red}{i^2=−1}\) and \(\textcolor{red}{i=\sqrt{−1}}\)

    The sum of a real number and an imaginary number is called a complex number. Examples of complex numbers are \(5+4i\) and \(3−2i\). All complex numbers can be written in the form a+bi where a and b are real numbers. Two important points:

    • The set of real numbers is a subset of the set of complex numbers where \(b=0\). Examples of real numbers are 2,7,12,−4.2.
    • The set of imaginary numbers is a subset of the set of complex numbers where a=0. Examples of imaginary numbers are \(i\), \(−4i\), \(\sqrt{2}i\).

    This means that the set of complex numbers includes real numbers, imaginary numbers, and combinations of real and imaginary numbers.

    When a quadratic function does not intersect the x-axis, it has complex roots. When solving for the roots of a function algebraically using the quadratic formula, you will end up with a negative under the square root symbol. With your knowledge of complex numbers, you can still state the complex roots of a function just like you would state the real roots of a function.

    Let's solve the quadratic equation: \(m^2−2m+5=0\)

    You can use the quadratic formula to solve. For this quadratic equation, \(a=1\), \(b=−2\), \(c=5\).

    \(\begin{array}{l}
    m=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
    m=\dfrac{-(\textcolor{red}{-2}) \pm \sqrt{(\textcolor{red}{-2})^{2}-4(\textcolor{red}{1})(\textcolor{red}{5})}}{2(\textcolor{red}{1})} \\
    m=\dfrac{2 \pm \sqrt{4-20}}{2} \\
    m=\dfrac{2 \pm \sqrt{-16}}{2} \quad \sqrt{-16}=\sqrt{16} \times i=4 i \\
    m=\dfrac{2 \pm 4 i}{2} \\
    m=1 \pm 2 i \\
    m=1+2 i \text { or } m=1-2 i
    \end{array}\)

    There are no real solutions to the equation. The solutions to the quadratic equation are \(1+2i\) and \(1−2i\).

    Now, let's solve the following equation by rewriting it as a quadratic and using the quadratic formula:

    \(\dfrac{3}{e+3}−\dfrac{2}{e+2}=1\)

    To rewrite as a quadratic equation, multiply each term by \((e+3)(e+2)\).

    \(\begin{aligned} \dfrac{3}{e+3} \textcolor{red}{(e+3)(e+2)}−2e+2\textcolor{red}{(e+3)(e+2)}=1\textcolor{red}{(e+3)(e+2)} \\ 3(e+2)−2(e+3)=(e+3)(e+2) \end{aligned}\)

    Expand and simplify.

    \(\begin{aligned} 3e+6−2e−6=e^2+2e+3e+6 \\ e^2+4e+6=0 \end{aligned}\)

    Solve using the quadratic formula. For this quadratic equation, a=1,b=4,c=6.

    \(\begin{array}{l}
    e=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
    e=\dfrac{-(\textcolor{red}{4}) \pm \sqrt{(\textcolor{red}{4})^{2}-4(\textcolor{red}{1})(\textcolor{red}{6})}}{2(\textcolor{red}{1})} \\
    e=\dfrac{-4 \pm \sqrt{16-24}}{2} \\
    e=\dfrac{-4 \pm \sqrt{-8}}{2} \quad \sqrt{-8}=\sqrt{8} \times i=\sqrt{4 \cdot 2} \times i=2 i \sqrt{2} \\
    e=\dfrac{-4 \pm 2 i \sqrt{2}}{2} \\
    e=-2 \pm i \sqrt{2} \\
    e=-2+i \sqrt{2} \text { or } e=-2-i \sqrt{2}
    \end{array}\)

    There are no real solutions to the equation. The solutions to the equation are −2+i\sqrt{2} and −2−i\sqrt{2}

    Finally, let's sketch the graph of the following quadratic function. What are the roots of this function?

    \(y=x^2−4x+5\)

    Use your calculator or a table to make a sketch of the function. You should get the following: 

    f-d_caa4a3888e16f0bf6d022408696eb42c4f9613bb6b03fd468a1213f4+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{2}\)

    As you can see, the quadratic function has no x-intercepts; therefore, the function has no real roots. To find the roots (which will be complex), you must use the quadratic formula.

    For this quadratic function, \(a=1\), \(b=−4\), \(c=5\).

    \(\begin{array}{l}
    x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
    x=\dfrac{-(\textcolor{red}{-4}) \pm \sqrt{(\textcolor{red}{-4})^{2}-4\textcolor{red}{1})(\textcolor{red}{5})}}{2(\textcolor{red}{1})} \\
    x=\dfrac{4 \pm \sqrt{16-20}}{2} \\
    x=\dfrac{4 \pm \sqrt{-4}}{2} \quad \sqrt{-4}=\sqrt{4} \times i=2 i \\
    x=\dfrac{4 \pm 2 i}{2} \\
    x=2 \pm i \\
    x=2+i \text { or } x=2-i
    \end{array}\)

    The complex roots of the quadratic function are 2+i and 2−i.

    Example \(\PageIndex{1}\)

    Earlier, you were asked to find the complex roots of \(y=x^2−2x+3\).

    Solution

    To find the complex roots of the function \(y=x^2−2x+3\), you must use the quadratic formula.

    For this quadratic function, \(a=1\), \(b=−2\), \(c=3\).

    \(\begin{array}{l}
    x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
    x=\dfrac{-(\textcolor{red}{-2}) \pm \sqrt{(\textcolor{red}{-2})^{2}-4(\textcolor{red}{1})(\textcolor{red}{3})}}{2(\textcolor{red}{1})} \\
    x=\dfrac{2 \pm \sqrt{4-12}}{2} \\
    x=\dfrac{2 \pm \sqrt{-8}}{2} \quad \sqrt{-8}=\sqrt{8} \times i=2 \sqrt{2} i \\
    x=\dfrac{2 \pm 2 \sqrt{2} i}{2} \\
    x=1 \pm \sqrt{2} i
    \end{array}\)

    Example \(\PageIndex{2}\)

    Solve the following quadratic equation. Express all solutions in simplest radical form.

    \(2n^2+n=−4\)

    Solution

    \(2n^2+n=−4\)

    Set the equation equal to zero.

    \(2n^2+n+4=0\)

    Solve using the quadratic formula.

    \(\begin{array}{l}
    x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
    n=\dfrac{-(\textcolor{red}{1}) \pm \sqrt{(\textcolor{red}{1})^{2}-4(\textcolor{red}{2})(\textcolor{red}{4})}}{2(\textcolor{red}{2})} \\
    n=\dfrac{-1 \pm \sqrt{1-32}}{4} \\
    n=\dfrac{-1 \pm \sqrt{-31}}{4} \\
    n=\dfrac{-1 \pm i \sqrt{31}}{4}
    \end{array}\)

    Example \(\PageIndex{3}\)

    Solve the following quadratic equation. Express all solutions in simplest radical form.

    \(m^2+(m+1)^2+(m+2)^2=−1\)

    Solution

    \(m^2+(m+1)^2+(m+2)^2=−1\)

    Expand and simplify.

    \(\begin{aligned} m^2+(m+1)(m+1)+(m+2)(m+2)&=−1 \\ m^2+m^2+m+m+1+m^2+2m+2m+4&=−1 \\ 3m^2+6m+5&=−1 \end{aligned}\)

    Write the equation in general form.

    \(3m^2+6m+6=0\)

    Divide by 3 to simplify the equation.

    \(\begin{aligned} \dfrac{3m^2}{\textcolor{red}{3}}+\dfrac{6m}{\textcolor{red}{3}}+\dfrac{6}{\textcolor{red}{3}}&=\dfrac{0}{\textcolor{red}{3}} \\ m^2+2m+2&=0 \end{aligned}\)

    Solve using the quadratic formula:

    \(\begin{array}{l}
    m=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
    m=\dfrac{-(\textcolor{red}{2}) \pm \sqrt{(\textcolor{red}{2})^{2}-4(\textcolor{red}{1})(\textcolor{red}{2})}}{2(\textcolor{red}{1})} \\
    m=\dfrac{-2 \pm \sqrt{4-8}}{2} \\
    m=\dfrac{-2 \pm \sqrt{-4}}{2} \\
    m=\dfrac{-2 \pm 2 i}{2} \\
    m=-1 \pm i
    \end{array}\)

    Example \(\PageIndex{4}\)

    Is it possible for a quadratic function to have exactly one complex root?

    Solution

    No, even in higher degree polynomials, complex roots will always come in pairs. Consider when you use the quadratic formula-- if you have a negative under the square root symbol, both the + version and the - version of the two answers will end up being complex.

    Review

    1. If a quadratic function has 2 x-intercepts, how many complex roots does it have? Explain.
    2. If a quadratic function has no x-intercepts, how many complex roots does it have? Explain.
    3. If a quadratic function has 1 x-intercept, how many complex roots does it have? Explain.
    4. If you want to know whether a function has complex roots, which part of the quadratic formula is it important to focus on?
    5. You solve a quadratic equation and get 2 complex solutions. How can you check your solutions?
    6. In general, you can attempt to solve a quadratic equation by graphing, factoring, completing the square, or using the quadratic formula. If a quadratic equation has complex solutions, what methods do you have for solving the equation?

    Solve the following quadratic equations. Express all solutions in simplest radical form.

    1. \(x^2+x+1=0\)
    2. \(5y^2−8y=−6\)
    3. \(2m^2−12m+19=0\)
    4. \(−3x^2−2x=2\)
    5. \(2x^2+4x=−11\)
    6. \(−x^2+x−23=0\)
    7. \(−3x^2+2x=14\)
    8. \(x^2+5=−x\)
    9. \(\dfrac{1}{2}d^2+4d=−12\)

    Review (Answers)

    To see the Review answers, open this PDF file and look for section 9.7. 

    Vocabulary

    Term Definition
    complex number A complex number is the sum of a real number and an imaginary number, written in the form \(a+bi\).
    complex root A complex root is a complex number that, when used as an input (\(x\)) value of a function, results in an output (\(y\)) value of zero.
    Imaginary Numbers An imaginary number is a number that can be written as the product of a real number and i.
    Quadratic Formula The quadratic formula states that for any quadratic equation in the form \(ax^2+bx+c=0\), \(x=\dfrac{−b\pm \sqrt{b^2−4ac}}{2a}\).
    Real Number A real number is a number that can be plotted on a number line. Real numbers include all rational and irrational numbers.

    Additional Resources

    Interactive Element

     

    Video: Using the Quadratic Formula

    Practice: Quadratic Formula and Complex Sums

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