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Write the number as a pure imaginary number. $$\sqrt{-\frac{4}{25}}$$

Short Answer

Expert verified
\(\sqrt{-\frac{4}{25}}\) as a pure imaginary number is \(\frac{2}{5}i\).

Step by step solution

01

Identify the Negative Sign

Start by identifying the negative sign of the square root. Recall that the square root of a negative number can be written in terms of \(i\), where \(i^2 = -1\). Thus, the expression can be rewritten as \(i \cdot \sqrt{\frac{4}{25}}\).
02

Simplify the Square Root

Take the square root of the positive fraction separately. The square root of \(\frac{4}{25}\) simplifies to \(\frac{2}{5}\).
03

Combine Results

We can now write the original expression as an imaginary number by combining the numerical result with \(i\), which gives \(\frac{2}{5}i\). Therefore, \(\sqrt{-\frac{4}{25}}\) in the form of a pure imaginary number is \(\frac{2}{5}i\).

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Key Concepts

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

Understanding Square Roots of Negative Numbers
Dealing with the square root of negative numbers might seem perplexing at first, as traditionally, square roots are thought to be applicable only to non-negative numbers. However, in mathematics, we overcome this hurdle using imaginary numbers.

Technically, you cannot have a real number that, when multiplied by itself, results in a negative number. But the concept of imaginary numbers addresses this very issue by introducing the imaginary unit, denoted as 'i', where the rule is defined as \(i^2 = -1\). Therefore, to find the square root of a negative number, you initially factor out the '-1' and treat it as \(i^2\), and then proceed to find the square root of the remaining positive part. For example, \(\sqrt{-4}\) becomes \(\sqrt{4} \times \sqrt{-1}\) or 2i, since \(\sqrt{-1}\) equals to 'i'.

This method allows us to extend the concept of square roots to negative numbers, enabling complex mathematical calculations and exploration of topics that wouldn't be possible with real numbers alone.
The Imaginary Unit i
At the core of simplifying square roots of negative numbers lies the imaginary unit, denoted as 'i'. The definition of 'i' is a simple yet pivotal concept: \(i = \sqrt{-1}\). Thus, 'i' provides a solution to square roots of negative one and by extension, to all negative numbers, when using a combination of 'i' and the square root of the positive counterpart.

The existence of 'i' opens up a whole new dimension in mathematics, known as complex numbers, which consist of a real part and an imaginary part. It's important to note that the imaginary unit isn't a real number and doesn't exist on the number line we use for real numbers. Instead, it exists in a plane known as the complex plane.

Furthermore, 'i' operates under specific rules, the most fundamental being \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and so on, repeating in a cycle. These properties are essential when performing arithmetic operations with complex numbers or simplifying expressions involving 'i'.
Simplifying Square Roots
When simplifying square roots, whether they involve imaginary numbers or not, we follow a systematic process to render the expression into its simplest form. This entails finding the prime factors of the number under the square root and pairing them off. Any number that can be paired is moved outside of the root as its square root, and those that cannot are left under the root sign.

For example, to simplify \(\sqrt{50}\), we would break down 50 into its prime factors: 2 and 25, which in turn is 5 squared. The square root of 25 (or 5 squared) comes out of the square root as a 5, while the 2 remains under the root, resulting in 5\(\sqrt{2}\).

However, when dealing with a fraction under a square root, we can simplify the square root of the numerator and denominator separately to make the process easier. If the fraction's numerator and denominator are both perfect squares, they simplify completely, otherwise, they may still have square roots remaining after simplification.

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Most popular questions from this chapter

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When designing buildings, engineers must pay careful attention to how different factors affect the load a structure can bear. The following table gives the load in terms of the weight of concrete that can be borne when threaded rod anchors of various diameters are used to form joints. $$\begin{array}{cc} \text { Diameter (in.) } & \text { Load (1b) } \\ \hline 0.3750 & 2105 \\ 0.5000 & 3750 \\ 0.6250 & 5875 \\ 0.7500 & 8460 \\ 0.8750 & 11,500 \end{array}$$ (a) Examine the table and explain why the relationship between the diameter and the load is not linear. (b) The function $$f(x)=14,926 x^{2}+148 x-51$$ gives the load (in pounds of concrete) that can be borne when rod anchors of diameter \(x\) (in inches) are employed. Use this function to determine the load for an anchor with a diameter of 0.8 inch. (c) since the rods are drilled into the concrete, the manufacturer's specifications sheet gives the load in terms of the diameter of the drill bit. This diameter is always 0.125 inch larger than the diameter of the anchor. Write the function in part (b) in terms of the diameter of the drill bit. The loads for the drill bits will be the same as the loads for the corresponding anchors. (Hint: Examine the table of values and see if you can present the table in terms of the diameter of the drill bit.)

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