Problem 35 - Intermediate Algebra

Question

Write each number in the form a \(+b i.\) a. \(-6-\sqrt{-9}\) b. \(3+\sqrt{-6}\)

Step-by-Step Solution

Answer

a. \\( -6 - 3i \\); b. \\( 3 + \sqrt{6} \, i \\).

Step 1: Identify the Square Root of Negative Numbers

Remember that the square root of a negative number involves the imaginary unit 'i', where \( i = \sqrt{-1} \). Thus, \( \sqrt{-x} = i\sqrt{x}\) for any positive number \(x\).

Step 2: Solve Part a: \(-6-\sqrt{-9}\)

In \(\sqrt{-9}\), rewrite \(-9\) as \(9 \cdot (-1)\). This gives \(\sqrt{9}\cdot\sqrt{-1} = 3i\). So, \( -6-\sqrt{-9} \) becomes \(-6 - 3i\).

Step 3: Solve Part b: \(3+\sqrt{-6}\)

Start with \(\sqrt{-6}\) and rewrite as \(\sqrt{6}\cdot\sqrt{-1} = \sqrt{6} \, i\). Thus, \(3+\sqrt{-6}\) becomes \(3 + \sqrt{6} \, i\).

Step 4: Express in \(a + b i\) Format

For part (a), the expression is already \( -6 - 3i \), which is in the form of \( a + bi \) where \(a = -6\) and \(b = -3\). In part (b), the expression is \(3 + \sqrt{6} \, i\), which is in the form of \( a + bi \) with \(a = 3\) and \(b = \sqrt{6}\).

Key Concepts

Imaginary Numbers

Imaginary numbers might sound complicated, but they're just a special kind of mathematical number used to describe what happens when we take the square root of a negative number. Usually, square roots only work with non-negative numbers because you can't easily find a number that multiplies by itself to give a negative result. That's where imaginary numbers come to the rescue! You can think of them as a creative way to work with these tricky square roots.

In the world of imaginary numbers, we introduce an imaginary unit called 'i'. So when you see an equation like \( \sqrt{-1} \), it equals 'i'. Any time you are solving a problem and you encounter the square root of a negative number, you'll use 'i'. For example, \( \sqrt{-9} \) translates to \( 3i \) because it's the same as \( 3 \times \sqrt{-1} \). This makes imaginary numbers an important part of complex numbers, which combine real and imaginary parts.

Square Roots

Square roots are a fundamental concept in math that involves finding a number which, when multiplied by itself, gives the original number. In other words, the square root of 9 is 3 because \( 3 \times 3 = 9 \). But what if you need to find the square root of a negative number? Math gets a bit creative here because such numbers do not exist on the real number line as we know it.

This is why imaginary numbers and the imaginary unit 'i' are so valuable. When dealing with negative numbers inside a square root, such as \( \sqrt{-9} \), you can think of it as \( \sqrt{9} \times \sqrt{-1} \). The \( \sqrt{9} \) is real and equals 3, while \( \sqrt{-1} \) is 'i'. So, combining these gives you the imaginary number \( 3i \). This method helps simplify calculations and ensures that sending negative numbers under the square root sign makes sense.

Imaginary Unit

The imaginary unit 'i' is the cornerstone of working with imaginary numbers. It is defined as \( i = \sqrt{-1} \). By introducing 'i', math finds a way to handle the seemingly impossible task of dealing with the square roots of negative numbers.

Here are some useful facts about 'i':

• \( i^2 = -1 \): Squaring 'i' gives -1, the very essence of its definition.
• \( i^3 = -i \): Which comes from multiplying \( i^2 \) by 'i'.
• \( i^4 = 1 \): Get back to 1 because multiplying \( i^3 \) by 'i' again reverts to one full rotation.

These properties make 'i' useful, especially in expressing complex numbers in the form \( a + bi \). A complex number is made up of a real part \( a \) and an imaginary part \( bi \). They are integral in fields including engineering, physics, and many areas of mathematics, where traditional number systems fall short.