Simplify with negative radicands in terms of $i$:



$x = \frac{5 + \sqrt{-49}}{6}$


A$-\frac{1}{3i}$
B$2i$
C$\frac{5}{6} - \frac{7i}{6}$
D$\frac{5}{6} + \frac{7i}{6}$

Verified Answer :

  • Identify the imaginary component by simplifying the square root of the negative number: \(\sqrt{-49} = \sqrt{49} \cdot \sqrt{-1} = 7i\).
  • Substitute the simplified term into the original equation: \(x = \frac{5+7i}{6}\).
  • Split the fraction into its real and imaginary parts to reach standard form: \(x = \frac{5}{6} + \frac{7i}{6}\).
  • Compare with the given options to find the correct choice: \(\boxed{d}\).

Explanation

  1. Problem Analysis

The problem asks to simplify the complex expression \(x = \frac{5 + \sqrt{-49}}{6}\) and express it in terms of the imaginary unit \(i\), where \(i = \sqrt{-1}\). A complex number is typically written in the standard form \(a + bi\), where \(a\) and \(b\) are real numbers. We are given four options to choose from, which helps guide the final format of our result.

  1. Simplifying the Radicand

The first step is to handle the term involving the negative square root, \(\sqrt{-49}\). We can rewrite the negative number inside the radical as the product of a positive number and -1:
\[\sqrt{-49} = \sqrt{49 \cdot (-1)}\]
Using the property of square roots \(\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}\) for non-negative \(a\), we get:
\[\sqrt{49 \cdot (-1)} = \sqrt{49} \cdot \sqrt{-1}\]
Since \(\sqrt{49} = 7\) and \(\sqrt{-1} = i\), the expression simplifies to:
\[\sqrt{-49} = 7i\]

  1. Final Simplification

Now, we substitute this back into our original expression for \(x\):
\[x = \frac{5 + 7i}{6}\]
To reach the standard complex form \(a + bi\), we divide both terms in the numerator by the denominator 6:
\[x = \frac{5}{6} + \frac{7i}{6}\]
Alternatively, this can be written as \(\frac{5}{6} + \frac{7}{6}i\). Looking at the options provided, this result matches option (d).

Examples

Complex numbers like the one we just simplified are essential in electrical engineering for analyzing alternating current (AC) circuits. They allow engineers to easily represent both magnitude and phase shifts of voltage and current using a single notation. For instance, when designing audio equipment or power grids, engineers calculate impedances using these complex values to ensure systems function correctly without overheating or signal loss. Understanding how to simplify these expressions is the first step toward mastering these complex engineering calculations.