Problem 99 Some equations have nonreal comp... [FREE SOLUTION]

Chapter 10: Problem 99

Some equations have nonreal complex solutions. To see how these nonreal complex solutions are related. Consider the equation \(x^{2}-2 x+26=0 .\) Show that each number is a solution of the equation. (a) \(1+5 i\) (b) \(1-5 i\)

Short Answer

Expert verified

Both \( 1+5i \) and \( 1-5i \) are solutions of the equation.

Step by step solution

Identify the Quadratic Equation

The given quadratic equation is: \[ x^{2} - 2x + 26 = 0 \]

Substitute First Solution

Substitute \( x = 1 + 5i \) into the equation: \[ (1+5i)^{2} - 2(1+5i) + 26 = 0 \]

Expand and Simplify

First calculate \((1+5i)^2 \): \[ (1+5i)^2 = 1^2 + 2(1)(5i) + (5i)^2 = 1 + 10i - 25 = -24 + 10i \]

Continue Simplification

Substitute \( -24 + 10i \) back into the equation: \[ -24 + 10i - 2(1 + 5i) + 26 = 0 \] Now simplify \[ -24 + 10i - 2 - 10i + 26 = 0 \] \[ -26 + 26 = 0 \] \[ 0 = 0 \] Therefore, \( 1 + 5i \) is a solution.

Substitute Second Solution

Substitute \( x = 1 - 5i \) into the equation: \[ (1-5i)^{2} - 2(1-5i) + 26 = 0 \]

Expand and Simplify

First calculate \((1-5i)^2 \): \[ (1-5i)^2 = 1^2 + 2(1)(-5i) + (-5i)^2 = 1 - 10i - 25 = -24 - 10i \]

Continue Simplification

Substitute \( -24 - 10i \) back into the equation: \[ -24 - 10i - 2(1 - 5i) + 26 = 0 \] Now simplify \[ -24 - 10i - 2 + 10i + 26 = 0 \] \[ -26 + 26 = 0 \] \[ 0 = 0 \] Therefore, \( 1 - 5i \) is a solution.

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Key Concepts

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

Quadratic Equation

A quadratic equation is a type of polynomial equation with the highest power of the unknown variable, typically denoted as x, being 2. It is generally written in the standard form: \[ ax^2 + bx + c = 0 \]where a, b, and c are constants, and a 鈮� 0. Quadratic equations are key in Algebra and appear in various real-world scenarios, from physics (projectile motion) to finance (parabolas for profit maximization). They can be solved using multiple methods:

Factoring: Expressing the quadratic equation as a product of binomials, if possible.
Quadratic Formula: Using the formula \[ x = \frac{-b \, \textpm \sqrt{b^2 - 4ac}}{2a} \] to find the roots directly.
Completing the Square: Transforming the quadratic into a perfect square form \[ (x + p)^2 = q \] to solve for x.
Graphing: Plotting the quadratic function and identifying the x-intercepts as the solutions.

In our example, the quadratic equation is:\[ x^2 - 2x + 26 = 0 \].We need to find the solutions to this equation, which lead us to the exploration of imaginary numbers and complex solutions.

Imaginary Numbers

To understand complex solutions, let's introduce imaginary numbers. An imaginary number is based on the imaginary unit i, where \[ i^2 = -1 \]. By this definition, the square root of a negative number is no longer undefined but can be expressed using i:

For example, \[ \sqrt{-25} = \sqrt{25} \cdot \sqrt{-1} = 5i \].

Imaginary numbers extend our number system to complex numbers, which have a real part and an imaginary part. In general, a complex number is written as:\[ a + bi \]where a and b are real numbers. The number a is the real part, and bi is the imaginary part. This framework is essential for solving certain quadratic equations that do not have real-number solutions.

Complex Solutions

When solving a quadratic equation, we sometimes encounter nonreal complex solutions, especially when the discriminant \[ b^2 - 4ac \] is negative. In such cases, the roots of the quadratic equation are complex numbers.In our example: \[x^2 - 2x + 26 = 0\]. Let's break this down with our given solutions:

First Solution: Substitute \[x = 1 + 5i \] into the equation:\[ (1+5i)^2 - 2(1+5i) + 26 = 0 \]Expand and simplify:
Second Solution: Substitute \[x = 1 - 5i \]:\[ (1-5i)^2 - 2(1-5i) + 26 = 0 \]Calculate \[(1-5i)^2\] and simplify:\[ (1-5i)^2 = 1 - 10i - 25 = -24 - 10i \]Substitute back into the equation:\[ -24 - 10i - 2(1 - 5i) + 26 = 0 \]Simplify further:\[ -24 - 10i - 2 + 10i + 26 = 0 \], which results in:\[ -26 + 26 = 0 \], verifying that \[1 - 5i\] is also a solution.

Complex solutions like these are crucial in expanding our understanding of quadratic equations beyond real numbers, thus opening up a broader and richer mathematical landscape.

91影视

Some equations have nonreal complex solutions. To see how these nonreal complex solutions are related. Consider the equation \(x^{2}-2 x+26=0 .\) Show that each number is a solution of the equation. (a) \(1+5 i\) (b) \(1-5 i\)

Short Answer

Step by step solution

Identify the Quadratic Equation

Substitute First Solution

Expand and Simplify

Continue Simplification

Substitute Second Solution

Expand and Simplify

Continue Simplification

Key Concepts

Quadratic Equation

Imaginary Numbers

Complex Solutions

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