Use the method described in Exercises 83–86, if applicable, and properties of absolute value to solve each equation or inequality. (Hint: Exercises 99 and 100 can be solved by inspection.) | 3x2 - 14x | = 5
Ch. 1 - Equations and Inequalities

Chapter 2, Problem 89a
Simplify each power of i. i25
Verified step by step guidance1
Recall that the imaginary unit \(i\) is defined such that \(i^2 = -1\).
Recognize that powers of \(i\) repeat in a cycle of 4: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\).
To simplify \(i^{25}\), find the remainder when 25 is divided by 4, since the powers repeat every 4.
Calculate \(25 \div 4\) which gives a quotient of 6 and a remainder of 1, so \(i^{25} = i^{4 \cdot 6 + 1} = (i^4)^6 \cdot i^1\).
Use the fact that \((i^4)^6 = 1^6 = 1\), so \(i^{25} = 1 \cdot i = i\).

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Imaginary Unit (i)
The imaginary unit i is defined as the square root of -1, satisfying i² = -1. It is the fundamental unit used to extend the real number system to complex numbers, allowing for the representation of numbers involving the square roots of negative values.
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Powers of i and Their Cyclic Pattern
Powers of i repeat in a cycle of four: i¹ = i, i² = -1, i³ = -i, and i⁴ = 1. This pattern repeats for higher powers, so simplifying i raised to any integer power involves finding the remainder when the exponent is divided by 4.
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Modular Arithmetic for Exponent Simplification
Modular arithmetic helps simplify powers by reducing the exponent modulo 4 in this context. For example, to simplify i^25, compute 25 mod 4 = 1, so i^25 = i¹ = i. This technique streamlines calculations involving cyclic patterns.
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Arithmetic Sequences - General Formula
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