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6.12: Finding Imaginary Solutions

Difficulty Level: At Grade Created by: CK-12
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Louis calculates that the area of a rectangle is represented by the equation 3x4+7x2=2. Did is calculate right? Explain.

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 f(x)=3x4x214. (#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)000=3x4x214=3x47x2+6x214=x2(3x27)+2(3x27)=(x2+2)(3x27)

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

x2+2=0x2=2andx=±2 or ±i23x27=03x2=7x2=73x=±73 or ±213

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)=x4+21x2+90.

Solution: When graphed, this function does not touch the xaxis. 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)0=x4+21x2+90=(x2+6)(x2+15)

Now, set each factor equal to zero and solve.

x2+6=0x2=6andx=±i6x2+15=0x2=15x=±i15

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 4i 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, 2x418x3+38x2+62x204 would have these roots as well.

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

3x4+7x2=2=3x4+7x22=03x4+7x22=(3x2+1)(x2+2)=0

Solving for x we get

3x2+1=0x2=13andx=±i13x2+2=0x2=2x=±i2

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)=25x3120x2+81x4

2. f(x)=4x4+35x29

3. Find the equation of a function with roots 4, 2 and 1i.

Answers

1. First, graph the function.

Using the Rational Root Theorem, the possible realistic zeros could be 125, 1 or 4. Let’s try these three possibilities using synthetic division.

Of these three possibilities, only 4 is a zero. The leftover polynomial, 25x220x+1 is not factorable, so we need to do the Quadratic Formula to find the last two zeros.

x=20±2024(25)(1)2(25)=20±40010050=20±10350 or 2±350.746 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)=4x4+35x29 is factorable. ac=36.

4x4+35x294x4+36x2x294x2(x2+9)1(x2+9)(x2+9)(4x21)

Setting each factor equal to zero, we have:

x2+9=0x2=9orx=±3i4x21=04x2=1x2=14x=±12

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, 2,2,1+i,1i. Multiply all 5 roots together.

(x4)(x2)(x+2)(x(1+i))(x(1i))(x4)(x22)(x22x+2)(x34x22x+8)(x22x+2)x56x4+8x34x220x+16

Practice

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

  1. f(x)=x4+x312x210x+20
  2. f(x)=4x320x23x+15
  3. f(x)=2x47x230
  4. f(x)=x3+5x2+12x+18
  5. f(x)=4x4+4x322x28x+40
  6. f(x)=3x4+4x215
  7. f(x)=2x36x2+9x27
  8. f(x)=6x47x3280x2419x+280
  9. f(x)=9x4+6x328x2+2x+11
  10. f(x)=2x519x4+30x3+97x220x+150

Find a function with the following roots.

  1. 4,i
  2. 3,2i
  3. 5,1+i
  4. 2,13,42
  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)=x5+x3+8x2+8.

Vocabulary

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.

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Date Created:
Mar 12, 2013
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