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Chapter 8: Problem 89

Complex numbers will appear again in this book in Chapter 9, when we study quadratic equations. The following exercises examine how a complex number can be a solution of a quadratic equation. Show that \(1+5 i\) is a solution of $$ x^{2}-2 x+26=0 $$ Then show that its conjugate is also a solution.

Short Answer

Expert verified
Both 1+5i and its conjugate 1-5i are solutions.

Step by step solution

01

- Substitute the Complex Number

Substitute the given complex number into the quadratic equation. Let鈥檚 start with the number 1+5i. Replace x with 1+5i in the equation: x^2 - 2x + 26.
02

- Calculate the Square

Calculate x^2 where x = 1 + 5i: (1 + 5i)^2 = 1^2 + 2(1)(5i) + (5i)^2.
03

- Simplify the Square

Simplify the expression: 1 + 10i + 25i^2. Since i^2 = -1, (1 + 10i + 25(-1)) = 1 + 10i - 25 = -24 + 10i.
04

- Substitute and Simplify

Substitute x = 1 + 5i in -2x: -2(1 + 5i) = -2 - 10i. Then substitute these into the quadratic equation: x^2 - 2x + 26 = (-24 + 10i) - (2 + 10i) + 26.
05

- Combine Like Terms

Combine the terms: (-24 - 2 + 26) + (10i - 10i). These simplify to 0: -26 + 26 =0, and 10i - 10i = 0. So 0 + 0 = 0.
06

- Show Solution for Conjugate

Now substitute the conjugate (1 - 5i) into the equation: (1 - 5i)^2 - 2(1 - 5i) + 26.
07

- Conjugate Square Calculation

Calculate (1 - 5i)^2. This gives: 1 - 10i + 25i^2, and i^2 = -1. Resulting in 1 - 10i - 25 = -24 - 10i.
08

- Substitute in Conjugate Equation

Substitute x = 1 - 5i in -2x to get: -2(1 - 5i) = -2 + 10i. Then substitute into the equation: (-24 - 10i) - (2 - 10i) + 26.
09

- Combine Like Terms in Conjugate

Combine the terms for the conjugate: -24 - 2 + 26=0 and -10i + 10i=0. So 0 + 0 = 0.

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Key Concepts

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

quadratic equations
A quadratic equation is a polynomial equation of the second degree. It typically follows the form: ax^2 + bx + c = 0, where 'a', 'b', and 'c' are constants. These equations graph as parabolas and can have two real solutions, one real solution, or two complex solutions. Understanding how complex numbers can be solutions for quadratic equations is essential. For instance, in the equation x^2 - 2x + 26 = 0, we see that it has complex solutions as the discriminant (b^2 - 4ac) is negative: (-2)^2 - 4(1)(26) = 4 - 104 = -100. Thus, no real solutions exist, but complex solutions do.
conjugate pairs
A conjugate pair consists of two complex numbers in the form of a + bi and a - bi. These pairs are important as they often appear when dealing with polynomial equations with real coefficients. If one complex number is a solution to a quadratic equation, its conjugate will also be a solution. For example, if 1 + 5i solves x^2 - 2x + 26 = 0, then 1 - 5i must also solve it. This results from substituting both values back into the equation, simplifying it, and showing both yield zero.
algebraic solutions
Algebraic solutions involve manipulating equations to isolate the variable. For quadratic equations, we frequently rely on factoring, completing the square, or the quadratic formula. Here, we substituted x = 1 + 5i into the equation x^2 - 2x + 26 and simplified, showing every term combines to zero, confirming it's a solution. A similar process is repeated for its conjugate 1 - 5i, illustrating that algebra justifies these complex numbers as valid solutions.
imaginary numbers
Imaginary numbers are a critical component when dealing with complex numbers. They are defined by the imaginary unit i, where i^2 = -1. Imaginary numbers typically appear with a real part to form complex numbers like 2 + 3i. Solving quadratic equations with complex solutions requires you to understand how to handle these imaginary components in calculations. In our example, the term 25i^2 simplifies to -25 because i^2 = -1, significantly impacting the result as you simplify the equation.

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