<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are reading an older version of this FlexBook® textbook: CK-12 Algebra II with Trigonometry Concepts Go to the latest version.

# 6.12: Finding Imaginary Solutions

Difficulty Level: At Grade Created by: CK-12
0%
Progress
Practice Fundamental Theorem of Algebra
Progress
0%

Louis calculates that the area of a rectangle is represented by the equation $3x^4 + 7x^2 = 2$ . Did is calculate right? Explain.

### 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 $f(x)=3x^4-x^2-14$ . (#12 from the previous problem set.)

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.

$f(x) &= 3x^4-x^2-14\\0 &= 3x^4-7x^2+6x^2-14\\0 &= x^2(3x^2-7)+2(3x^2-7)\\0 &= (x^2+2)(3x^2-7)$

Now, because neither factor can be factored further and there is no $x-$ term, we can set each equal to zero and solve.

$& && 3x^2-7=0\\& x^2+2=0 && 3x^2=7\\& x^2=-2 \qquad \qquad \qquad \quad and && x^2=\frac{7}{3}\\& x=\pm \sqrt{-2} \ or \ \pm i \sqrt{2} && x=\pm \sqrt{\frac{7}{3}} \ or \ \pm \frac{\sqrt{21}}{3}$

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 $g(x)=x^4+21x^2+90$ .

Solution: When graphed, this function does not touch the $x-$ axis. Therefore, all the solutions are imaginary. To solve, this function can be factored like a quadratic equation. The factors of 90 that add up to 21 are 6 and 15.

$g(x) &= x^4+21x^2+90\\0 &= (x^2+6)(x^2+15)$

Now, set each factor equal to zero and solve.

$& x^2+6=0 && x^2+15=0\\& x^2=-6 \qquad \qquad \qquad and && x^2=-15\\& x=\pm i \sqrt{6} && x=\pm i \sqrt{15}$

#### Example C

Find the function that has the solution 3, -2, and $4 + i$ .

Solution: Notice that one of the given solutions is imaginary. Imaginary solutions always come in pairs, so $4 - i$ 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, $2x^4-18x^3+38x^2+62x-204$ would have these roots as well.

Intro Problem Revisit First we need to change the equation to standard form. Then we can factor it.

$3x^4 + 7x^2 = 2\\= 3x^4 + 7x^2 - 2 = 0\\3x^4 + 7x^2 - 2 = (3x^2 + 1)(x^2 + 2) = 0$

Solving for x we get

$3x^2+1=0 && x^2+2=0\\& x^2=\frac{-1}{3} \qquad \qquad \qquad and && x^2=-2\\& x=\pm i \sqrt{\frac{1}{3}} && x=\pm i \sqrt{2}$

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. $f(x)=25x^3-120x^2+81x-4$

2. $f(x)=4x^4+35x^2-9$

3. Find the equation of a function with roots 4, $\sqrt{2}$ and $1-i$ .

1. First, graph the function.

Using the Rational Root Theorem, the possible realistic zeros could be $\frac{1}{25}$ , 1 or 4. Let’s try these three possibilities using synthetic division.

Of these three possibilities, only 4 is a zero. The leftover polynomial, $25x^2-20x+1$ is not factorable, so we need to do the Quadratic Formula to find the last two zeros.

$x &= \frac{20 \pm \sqrt{20^2-4(25)(1)}}{2(25)}\\&= \frac{20 \pm \sqrt{400-100}}{50}\\& =\frac{20 \pm 10 \sqrt{3}}{50} \ or \ \frac{2 \pm \sqrt{3}}{5} \approx 0.746 \ and \ 0.054$

$^*$ Helpful Hint: Always find the decimal values of each zero to make sure they match up with the graph.

2. $f(x)=4x^4+35x^2-9$ is factorable. $ac = -36$ .

$& 4x^4+35x^2-9\\& 4x^4+36x^2-x^2-9\\& 4x^2(x^2+9)-1(x^2+9)\\& (x^2+9)(4x^2-1)$

Setting each factor equal to zero, we have:

$& && 4x^2-1=0\\& x^2+9=0 && 4x^2=1\\& x^2=-9 \quad \qquad \qquad or && x^2=\frac{1}{4}\\& x=\pm 3i && x=\pm \frac{1}{2}$

$^*$ 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, $\sqrt{2}, {\color{red}-\sqrt{2}},1+i,{\color{red}1-i}$ . Multiply all 5 roots together.

$& (x-4)(x-\sqrt{2})(x+\sqrt{2})(x-(1+i))(x-(1-i))\\& (x-4)(x^2-2)(x^2-2x+2)\\& (x^3-4x^2-2x+8)(x^2-2x+2)\\& x^5-6x^4+8x^3-4x^2-20x+16$

### Practice

Find all solutions to the following functions. Use any method.

1. $f(x)=x^4+x^3-12x^2-10x+20$
2. $f(x)=4x^3-20x^2-3x+15$
3. $f(x)=2x^4-7x^2-30$
4. $f(x)=x^3+5x^2+12x+18$
5. $f(x)=4x^4+4x^3-22x^2-8x+40$
6. $f(x)=3x^4+4x^2-15$
7. $f(x)=2x^3-6x^2+9x-27$
8. $f(x)=6x^4-7x^3-280x^2-419x+280$
9. $f(x)=9x^4+6x^3-28x^2+2x+11$
10. $f(x)=2x^5-19x^4+30x^3+97x^2-20x+150$

Find a function with the following roots.

1. $4, i$
2. $-3, -2i$
3. $\sqrt{5}, -1 + i$
4. $2, \frac{1}{3}, 4-\sqrt{2}$
5. Writing Write down the steps you use to find all the zeros of a polynomial function.
6. Writing Why do imaginary and irrational roots always come in pairs?
7. Challenge Find all the solutions to $f(x)=x^5+x^3+8x^2+8$ .

### Vocabulary Language: English

Complex Conjugate

Complex Conjugate

Complex conjugates are pairs of complex binomials. The complex conjugate of $a+bi$ is $a-bi$. When complex conjugates are multiplied, the result is a single real number.
complex number

complex number

A complex number is the sum of a real number and an imaginary number, written in the form $a + bi$.
conjugate pairs theorem

conjugate pairs theorem

The conjugate pairs theorem states that if $f(z)$ is a polynomial of degree $n$, with $n\ne0$ and with real coefficients, and if $f(z_{0})=0$, where $z_{0}=a+bi$, then $f(z_{0}^{*})=0$. Where $z_{0}^{*}$ is the complex conjugate of $z_{0}$.
fundamental theorem of algebra

fundamental theorem of algebra

The fundamental theorem of algebra states that if $f(x)$ is a polynomial of degree $n\ge 1$, then $f(x)$ has at least one zero in the complex number domain. In other words, there is at least one complex number $c$ such that $f(c)=0$. The theorem can also be stated as follows: an $n^{th}$ degree polynomial with real or complex coefficients has, with multiplicity, exactly $n$ complex roots.
Imaginary Number

Imaginary Number

An imaginary number is a number that can be written as the product of a real number and $i$.
Imaginary Numbers

Imaginary Numbers

An imaginary number is a number that can be written as the product of a real number and $i$.
Polynomial

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

Roots

The roots of a function are the values of x that make y equal to zero.
Zero

Zero

The zeros of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.
Zeroes

Zeroes

The zeroes of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.

Mar 12, 2013

Jun 04, 2015