Problem 51

Question

Evaluate the expression and write the result in the form \(a+b i .\) $$ \frac{\sqrt{-36}}{\sqrt{-2} \sqrt{-9}} $$

Step-by-Step Solution

Verified
Answer
The result is \(0 - \sqrt{2}i\).
1Step 1: Simplify the Numerator
The numerator of the expression is \(\sqrt{-36}\). We recognize \(-36\) as \(-1 \times 36\), and \(\sqrt{-1}\) is \(i\). Thus, \(\sqrt{-36} = \sqrt{36} \cdot \sqrt{-1} = 6i\).
2Step 2: Simplify the Denominator
The denominator consists of \(\sqrt{-2} \times \sqrt{-9}\). Using the property \(\sqrt{-x} = \sqrt{x} \cdot i\), we find \(\sqrt{-2} = \sqrt{2} \cdot i\) and \(\sqrt{-9} = \sqrt{9} \cdot i = 3i\). Therefore, the denominator is \(\sqrt{2} \cdot i \times 3i = 3 \sqrt{2} \cdot i^2\).
3Step 3: Simplify \(i^2\)
Recall that \(i^2 = -1\). Substituting this into the denominator gives \(3 \sqrt{2} \cdot (-1) = -3 \sqrt{2}\).
4Step 4: Write the Fraction
Now the expression becomes \(\frac{6i}{-3\sqrt{2}}\).
5Step 5: Simplify the Fraction
To simplify, divide both the numerator and the denominator by \(-3\). This gives: \(\frac{6i}{-3\sqrt{2}} = \frac{-2i}{\sqrt{2}}\).
6Step 6: Rationalize the Denominator
Multiply the numerator and the denominator by \(\sqrt{2}\) to rationalize the denominator: \(\frac{-2i}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{-2i \sqrt{2}}{2}\). Simplify to get \(-i \sqrt{2}\).
7Step 7: Write in Standard Form
Write the result of \(-i \sqrt{2}\) in the form \(a + bi \). Here \(a = 0\) and \(b = -\sqrt{2}\).
8Step 8: Final Result
The expression \(a + bi\) is \(0 - \sqrt{2}i\).

Key Concepts

Imaginary NumbersRationalizing the DenominatorSimplifying Radicals
Imaginary Numbers
Imaginary numbers are a fundamental part of complex numbers. They arise from the necessity to solve certain mathematical problems, such as the square root of negative numbers. Normally, square roots of negative numbers do not have real solutions. This is where the imaginary unit, denoted as \(i\), is introduced.
  • \(i\) is defined as \(\sqrt{-1}\). Therefore, \(i^2 = -1\).
  • Using this definition, any negative number can be expressed as the product of a positive number and \(-1\). For example, \(\sqrt{-36}\) can be broken down to \(\sqrt{36} \times \sqrt{-1} = 6i\).
  • The concept extends the number line into a new dimension, allowing for a complete set of solutions to polynomials.
The combination of real and imaginary numbers forms what we call complex numbers, expressed generally as \(a+bi\), where \(a\) and \(b\) are real numbers.
Rationalizing the Denominator
Rationalizing the denominator is a method used to eliminate radicals from the denominator of a fraction. It is important because having radicals in the denominator is not considered to be in its simplest form in mathematics.
  • To rationalize \(\frac{-2i}{\sqrt{2}}\), multiply both the numerator and denominator by \(\sqrt{2}\). This action leverages the property \(\sqrt{a} \times \sqrt{a} = a\).
  • This transforms the expression to \(\frac{-2i \sqrt{2}}{2}\). Notice that \(\sqrt{2} \times \sqrt{2} = 2\) neatly removes the radical from the denominator.
By performing this multiplication, the denominator becomes a rational number, and the fraction simplifies easily to \(-i \sqrt{2}\). Rationalizing makes expressions easier to work with, especially in further calculations or when combining fractions.
Simplifying Radicals
Simplifying radicals involves breaking down expressions containing square roots (or other roots) into their simplest form. It's a key skill in algebra that helps make expressions easier to manage and compare.
  • For a positive number under the square root, you look for perfect squares within the number. For example, \(\sqrt{36} = 6\) because 36 is a perfect square.
  • When dealing with negative numbers, the inclusion of \(i\) is required. For example, \(\sqrt{-9}\) becomes \(3i\) because \(\sqrt{9} = 3\) and \(\sqrt{-1} = i\).
After simplifying each radical, you can combine them if needed. Knowing how to simplify radicals is crucial for understanding complex numbers and is widely applicable in higher math involving roots, such as calculus and trigonometry.