6.12: Finding Imaginary Solutions
Louis calculates that the area of a rectangle is represented by the equation \begin{align*}3x^4 + 7x^2 = 2\end{align*}
Watch This
James Sousa: Ex 4: Find the Zeros of a Polynomial Function with Imaginary Zeros
Guidance
In #12 from the previous problem set, there are two imaginary solutions. Imaginary solutions always come in pairs. To find the imaginary solutions to a function, use the Quadratic Formula. If you need a little review on imaginary numbers and how to solve a quadratic equation with complex solutions see the Quadratic Equations chapter.
Example A
Solve \begin{align*}f(x)=3x^4-x^2-14\end{align*}
Solution: First, this quartic function can be factored just like a quadratic equation. See the Factoring Polynomials in Quadratic Form concept from this chapter for review.
\begin{align*}f(x) &= 3x^4-x^2-14\\
0 &= 3x^4-7x^2+6x^2-14\\
0 &= x^2(3x^2-7)+2(3x^2-7)\\
0 &= (x^2+2)(3x^2-7)\end{align*}
Now, because neither factor can be factored further and there is no \begin{align*}x-\end{align*}
\begin{align*}& && 3x^2-7=0\\
& x^2+2=0 && 3x^2=7\\
& x^2=-2 \qquad \qquad \qquad \quad and && x^2=\frac{7}{3}\\
& x=\pm \sqrt{-2} \ or \ \pm i \sqrt{2} && x=\pm \sqrt{\frac{7}{3}} \ or \ \pm \frac{\sqrt{21}}{3}\end{align*}
Including the imaginary solutions, there are four, which is what we would expect because the degree of this function is four.
Example B
Find all the solutions of the function \begin{align*}g(x)=x^4+21x^2+90\end{align*}
Solution: When graphed, this function does not touch the \begin{align*}x-\end{align*}
\begin{align*}g(x) &= x^4+21x^2+90\\
0 &= (x^2+6)(x^2+15)\end{align*}
Now, set each factor equal to zero and solve.
\begin{align*}& x^2+6=0 && x^2+15=0\\
& x^2=-6 \qquad \qquad \qquad and && x^2=-15\\
& x=\pm i \sqrt{6} && x=\pm i \sqrt{15}\end{align*}
Example C
Find the function that has the solution 3, -2, and \begin{align*}4 + i\end{align*}.
Solution: Notice that one of the given solutions is imaginary. Imaginary solutions always come in pairs, so \begin{align*}4 - i\end{align*} is also a factor, they are complex conjugates. Now, translate each solution into a factor and multiply them all together.
Any multiple of this function would also have these roots. For example, \begin{align*}2x^4-18x^3+38x^2+62x-204\end{align*} would have these roots as well.
Intro Problem Revisit First we need to change the equation to standard form. Then we can factor it.
\begin{align*}3x^4 + 7x^2 = 2\\ = 3x^4 + 7x^2 - 2 = 0\\ 3x^4 + 7x^2 - 2 = (3x^2 + 1)(x^2 + 2) = 0 \end{align*}
Solving for x we get
\begin{align*} 3x^2+1=0 && x^2+2=0\\ & x^2=\frac{-1}{3} \qquad \qquad \qquad and && x^2=-2\\ & x=\pm i \sqrt{\frac{1}{3}} && x=\pm i \sqrt{2}\end{align*}
All of the solutions are imaginary and the area of a rectangle must have real solutions. Therefore Louis did not calculate correctly.
Guided Practice
Find all the solutions to the following functions.
1. \begin{align*}f(x)=25x^3-120x^2+81x-4\end{align*}
2. \begin{align*}f(x)=4x^4+35x^2-9\end{align*}
3. Find the equation of a function with roots 4, \begin{align*}\sqrt{2}\end{align*} and \begin{align*}1-i\end{align*}.
Answers
1. First, graph the function.
Using the Rational Root Theorem, the possible realistic zeros could be \begin{align*}\frac{1}{25}\end{align*}, 1 or 4. Let’s try these three possibilities using synthetic division.
Of these three possibilities, only 4 is a zero. The leftover polynomial, \begin{align*}25x^2-20x+1\end{align*} is not factorable, so we need to do the Quadratic Formula to find the last two zeros.
\begin{align*}x &= \frac{20 \pm \sqrt{20^2-4(25)(1)}}{2(25)}\\ &= \frac{20 \pm \sqrt{400-100}}{50}\\ & =\frac{20 \pm 10 \sqrt{3}}{50} \ or \ \frac{2 \pm \sqrt{3}}{5} \approx 0.746 \ and \ 0.054\end{align*}
\begin{align*}^*\end{align*}Helpful Hint: Always find the decimal values of each zero to make sure they match up with the graph.
2. \begin{align*}f(x)=4x^4+35x^2-9\end{align*} is factorable. \begin{align*}ac = -36\end{align*}.
\begin{align*}& 4x^4+35x^2-9\\ & 4x^4+36x^2-x^2-9\\ & 4x^2(x^2+9)-1(x^2+9)\\ & (x^2+9)(4x^2-1)\end{align*}
Setting each factor equal to zero, we have:
\begin{align*}& && 4x^2-1=0\\ & x^2+9=0 && 4x^2=1\\ & x^2=-9 \quad \qquad \qquad or && x^2=\frac{1}{4}\\ & x=\pm 3i && x=\pm \frac{1}{2}\end{align*}
\begin{align*}^*\end{align*}This problem could have also been done by using the same method from #1.
3. Recall that irrational and imaginary roots come in pairs. Therefore, all the roots are 4, \begin{align*}\sqrt{2}, {\color{red}-\sqrt{2}},1+i,{\color{red}1-i}\end{align*}. Multiply all 5 roots together.
\begin{align*}& (x-4)(x-\sqrt{2})(x+\sqrt{2})(x-(1+i))(x-(1-i))\\ & (x-4)(x^2-2)(x^2-2x+2)\\ & (x^3-4x^2-2x+8)(x^2-2x+2)\\ & x^5-6x^4+8x^3-4x^2-20x+16\end{align*}
Practice
Find all solutions to the following functions. Use any method.
- \begin{align*}f(x)=x^4+x^3-12x^2-10x+20\end{align*}
- \begin{align*}f(x)=4x^3-20x^2-3x+15\end{align*}
- \begin{align*}f(x)=2x^4-7x^2-30\end{align*}
- \begin{align*}f(x)=x^3+5x^2+12x+18\end{align*}
- \begin{align*}f(x)=4x^4+4x^3-22x^2-8x+40\end{align*}
- \begin{align*}f(x)=3x^4+4x^2-15\end{align*}
- \begin{align*}f(x)=2x^3-6x^2+9x-27\end{align*}
- \begin{align*}f(x)=6x^4-7x^3-280x^2-419x+280\end{align*}
- \begin{align*}f(x)=9x^4+6x^3-28x^2+2x+11\end{align*}
- \begin{align*}f(x)=2x^5-19x^4+30x^3+97x^2-20x+150\end{align*}
Find a function with the following roots.
- \begin{align*}4, i\end{align*}
- \begin{align*}-3, -2i\end{align*}
- \begin{align*}\sqrt{5}, -1 + i\end{align*}
- \begin{align*}2, \frac{1}{3}, 4-\sqrt{2}\end{align*}
- Writing Write down the steps you use to find all the zeros of a polynomial function.
- Writing Why do imaginary and irrational roots always come in pairs?
- Challenge Find all the solutions to \begin{align*}f(x)=x^5+x^3+8x^2+8\end{align*}.
Complex Conjugate
Complex conjugates are pairs of complex binomials. The complex conjugate of is . When complex conjugates are multiplied, the result is a single real number.complex number
A complex number is the sum of a real number and an imaginary number, written in the form .conjugate pairs theorem
The conjugate pairs theorem states that if is a polynomial of degree , with and with real coefficients, and if , where , then . Where is the complex conjugate of .fundamental theorem of algebra
The fundamental theorem of algebra states that if is a polynomial of degree , then has at least one zero in the complex number domain. In other words, there is at least one complex number such that . The theorem can also be stated as follows: an degree polynomial with real or complex coefficients has, with multiplicity, exactly complex roots.Imaginary Number
An imaginary number is a number that can be written as the product of a real number and .Imaginary Numbers
An imaginary number is a number that can be written as the product of a real number and .Polynomial
A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.Roots
The roots of a function are the values of x that make y equal to zero.Zero
The zeros of a function are the values of that cause to be equal to zero.Zeroes
The zeroes of a function are the values of that cause to be equal to zero.Image Attributions
Here you'll find all the solutions to any polynomial, including imaginary solutions.