/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Find a polynomial equation with ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find a polynomial equation with real coefficients that has the given roots. $$-9 i, 9 i$$

Short Answer

Expert verified
\(x^2 + 81\)

Step by step solution

01

- Identifying the roots

The given roots of the polynomial are \(-9i\) and \(9i\). These are purely imaginary numbers.
02

- Using the fact about polynomial roots

Since the polynomial has real coefficients, the complex roots must occur in conjugate pairs. The roots given, \(-9i\) and \(9i\), already form such a pair.
03

- Constructing factors from roots

From the roots \(-9i\) and \(9i\), the factors of the polynomial are \((x + 9i)\) and \((x - 9i)\).
04

- Writing the polynomial as a product of factors

The polynomial can be written as \((x + 9i)(x - 9i)\).
05

- Expanding the polynomial

Expand the product using the difference of squares formula: \((x + 9i)(x - 9i) = x^2 - (9i)^2 = x^2 - 81(-1) = x^2 + 81\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Roots
In the realm of polynomial equations, complex roots often enter the scene. Complex numbers extend our traditional set of real numbers, adding in elements of the form \(a + bi\), where \(i\) represents the imaginary unit. This unit is defined by the property that \(i^2 = -1\). Complex numbers are crucial because they ensure that any polynomial equation—no matter its degree—has a solution.
When dealing with polynomial equations with real coefficients, you must remember that complex roots always come in pairs known as 'conjugate pairs'. Understanding this will help you accurately construct polynomials from given roots.
Conjugate Pairs
When a polynomial has real coefficients, any complex root must have a conjugate. A conjugate pair is made up of two complex numbers of the form \(a + bi\) and \(a - bi\). This is because the sum and product of conjugate pairs are both real numbers. Let's look at the roots in this exercise:
  • Root 1: \(-9i\)
  • Root 2: \(9i\)
These roots are purely imaginary and indeed form a conjugate pair because their imaginary parts are equal in magnitude but opposite in sign.
Constructing the polynomial from these roots requires creating factors from each root and then multiplying them together. For the roots \(-9i\) and \(9i\), the factors are \((x + 9i)\) and \((x - 9i)\).Multiplying these factors yields a polynomial with real coefficients, thanks to an important algebraic identity: the difference of squares.
Difference of Squares
The difference of squares formula is a handy tool when you're dealing with conjugate pairs. This algebraic identity is expressed as
\[a^2 - b^2 = (a - b)(a + b)\]
In our context, for the roots \(-9i\) and \(9i\), we use
\[(x + 9i)(x - 9i) = x^2 - (9i)^2\]
The imaginary unit \(i\) satisfies \(i^2 = -1\). Applying this to \(9i\), we have
\[9^2*i^2 = 81*(-1) = -81\]

Thus,
  • The factors \((x + 9i)\) and \((x - 9i)\) multiply to form \[x^2 - (9i)^2 = x^2 - (-81) = x^2 + 81\]
This resulting polynomial, \(x^2 + 81\), has real coefficients and the given complex roots. Mastering these concepts provides a strong foundation in understanding higher-level algebra and the behavior of polynomial equations.

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