Problem 52 - Precalculus Mathematics for Calculus

Question

Evaluate the radical expression and express the result in the form \(a+b i\) $$(\sqrt{3}-\sqrt{-4})(\sqrt{6}-\sqrt{-8})$$

Step-by-Step Solution

Answer

The result is \(-\sqrt{2} - 4\sqrt{6}i\).

Step 1: Simplify Each Square Root Expression

First, simplify the square root expressions individually. The expression \(\sqrt{-4}\) becomes \(2i\) because \(\sqrt{-4} = \sqrt{4} \cdot \sqrt{-1} = 2i\). Similarly, \(\sqrt{-8}\) simplifies to \(2\sqrt{2}i\) since \(\sqrt{-8} = \sqrt{4 \times 2} = 2\sqrt{2} \cdot \sqrt{-1} = 2\sqrt{2}i\).

Step 2: Write the Complex Expression

Substitute the simplified complex numbers back into the original expression: \((\sqrt{3} - 2i)(\sqrt{6} - 2\sqrt{2}i)\).

Step 3: Apply the Distributive Property

Use the distributive property (the binomial expansion):\((\sqrt{3})(\sqrt{6}) + (\sqrt{3})(-2\sqrt{2}i) + (-2i)(\sqrt{6}) + (-2i)(-2\sqrt{2}i)\). This gives us four products to calculate.

Step 4: Simplify Each Term

Calculate each term separately:1. \(\sqrt{3} \cdot \sqrt{6} = \sqrt{18} = 3\sqrt{2}\).2. \(\sqrt{3} \cdot -2\sqrt{2}i = -2\sqrt{6}i\).3. \(-2i \cdot \sqrt{6} = -2\sqrt{6}i\).4. \(-2i \cdot -2\sqrt{2}i = 4\sqrt{2}i^2 = 4\sqrt{2}(-1) = -4\sqrt{2}\).

Step 5: Combine Like Terms

Combine all four terms: \(3\sqrt{2} - 4\sqrt{2} + (-2\sqrt{6}i - 2\sqrt{6}i)\).- Combining real terms: \(3\sqrt{2} - 4\sqrt{2} = -\sqrt{2}\).- Combining imaginary terms: \(-2\sqrt{6}i - 2\sqrt{6}i = -4\sqrt{6}i\).

Step 6: Write the Expression in the Form \(a + bi\)

Arrange the expression so that it is in the standard complex form \(a + bi\):\(-\sqrt{2} - 4\sqrt{6}i\).

Key Concepts

Radical Expressions

Radical expressions often involve square roots, which can include both positive and negative numbers under the radical. Understanding how to handle negative radicands is crucial because it leads directly to the concept of complex numbers. A negative inside the square root means you deal with imaginary numbers. For instance, in the expression \(\sqrt{-4}\), the \(-4\) suggests a complex number. This expression simplifies to \(2i\) because \(\sqrt{4} = 2\) and \(\sqrt{-1} = i\). Similarly, \(\sqrt{-8}\) can be broken down to \(2\sqrt{2}i\), following the same process. Being comfortable with these transformations helps in managing expressions that contain imaginary components.

Complex Conjugate

In the world of complex numbers, the concept of a complex conjugate helps in simplifying expressions, especially during multiplication and division. A complex conjugate of a number \(a + bi\) is \(a - bi\). When these are multiplied, the imaginary parts cancel each other, resulting in a real number. This is a powerful tool when dealing with complex expressions. For example, multiplying complex conjugates \((a+bi)(a-bi)\) gives \(a^2 + b^2\), demonstrating why this property plays a significant role in finding real-number solutions to complex equations.
Understanding this property allows you to simplify and manipulate complex expressions efficiently.

Imaginary Numbers

Imaginary numbers form the basis for complex numbers. The imaginary unit, represented as \(i\), is defined such that \(i^2 = -1\). This property is fundamental in transforming negative square roots into complex numbers. For example, \(\sqrt{-4}\) becomes \(2i\), as we rewrite \(\sqrt{-4} = \sqrt{4} \cdot \sqrt{-1} = 2 \cdot i\). Imaginary numbers allow mathematicians to solve equations that do not have real solutions. They are also critical in many fields like engineering and physics, where real-world phenomena cannot be explained using only real numbers.
By understanding imaginary numbers, students unlock the potential to solve complex equations and better grasp advanced mathematical concepts.