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Chapter 2: Problem 65

Write the complex number in standard form.$$\sqrt{-6} \cdot \sqrt{-2}$$.

Short Answer

Expert verified
The standard form of the given complex number is \( -\sqrt{12} \)

Step by step solution

01

Identify the imaginary numbers

The given task is to multiply \( \sqrt{-6} \) and \( \sqrt{-2} \). Both roots signify imaginary numbers. We know that \( \sqrt{-1} = i \). So, \( \sqrt{-6} = \sqrt{6} \cdot \sqrt{-1} = \sqrt{6} \cdot i = i\sqrt{6} \) and \( \sqrt{-2} = \sqrt{2} \cdot \sqrt{-1} = \sqrt{2} \cdot i = i\sqrt{2} \).
02

Multiply the imaginary numbers

Now we will multiply these two imaginary numbers. \( i\sqrt{6} \cdot i\sqrt{2} = i^2 \sqrt{12} \). We know that \( i^2 = -1 \). Hence, we substitute \( -1 \) in place of \( i^2 \).
03

Simplify

So, \( i^2 \sqrt{12} = -1 \sqrt{12} = -\sqrt{12} \).

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Key Concepts

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

Imaginary Numbers
Imaginary numbers are numbers that give a negative result when squared. Their main purpose is to extend the numerical system beyond real numbers. Imaginary numbers are expressed using the symbol "i," which stands for the square root of -1. Therefore, if you encounter \( \sqrt{-1} \), it can be written as "i".
  • For any negative number inside a square root, you can extract the "i" and write the result as the product of "i" and the square root of the positive part.
  • For example, \( \sqrt{-6} \) can be rewritten as \( i\sqrt{6} \), following this principle.
Imaginary numbers are essential for solving many equations in mathematics that have no real solutions. When combined with real numbers, they form complex numbers.
Standard Form
The standard form of a complex number is a way of expressing it in the simplest and most recognized format. A complex number in standard form looks like \( a + bi \), where "a" is the real part, and "b" is the coefficient of the imaginary part. This format makes it easier to perform arithmetic operations, such as addition or subtraction, on complex numbers.
  • The real part "a" is a regular real number.
  • The imaginary part "bi" highlights the component that involves the imaginary unit "i".
For example, the expression resulting from our problem, \( -\sqrt{12} \), can be viewed as having a real part of \(-\sqrt{12}\) (since it doesn't visibly contain an "i" part). Sometimes, if needed, we can also express it as the sum \(-\sqrt{12} + 0i\) to clearly fit into the \( a + bi \) format.
Multiplication of Complex Numbers
Multiplying complex numbers involves distributing each part of one number with each part of the other number, similar to multiplying binomials. Let's break this process down using our given expression, \( i\sqrt{6} \cdot i\sqrt{2} \).
  • First, multiply the "i" terms: \( i \cdot i = i^2 = -1 \). This results from the definition of the imaginary unit.
  • Next, multiply the remaining square roots: \( \sqrt{6} \cdot \sqrt{2} = \sqrt{12} \).
Together, these become \( i^2 \sqrt{12} \), which simplifies to \( -1 \sqrt{12} = -\sqrt{12} \).
When performing this multiplication, it鈥檚 important to remember that \( i^2 \) always equals \(-1\), and this is crucial for simplifying the results correctly.

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Most popular questions from this chapter

Determine whether the statement is true or false. Justify your answer. The graph of a rational function can have a vertical asymptote, a horizontal asymptote, and a slant asymptote.

Write the quadratic function $$f(x)=a x^{2}+b x+c$$ in standard form to verify that the vertex occurs at $$\left(-\frac{b}{2 a}, f\left(-\frac{b}{2 a}\right)\right)$$

Find the rational zeros of the polynomial function. $$f(x)=x^{3}-\frac{3}{2} x^{2}-\frac{23}{2} x+6=\frac{1}{2}\left(2 x^{3}-3 x^{2}-23 x+12\right)$$

Use a graphing utility to graph the quadratic function. Identify the vertex, axis of symmetry, and \(x\) -intercept(s). Then check your results algebraically by writing the quadratic function in standard form. $$f(x)=\frac{3}{5}\left(x^{2}+6 x-5\right)$$

Use synthetic division to verify the upper and lower bounds of the real zeros of \(f\) \(f(x)=x^{3}-4 x^{2}+1\) (a) Upper: \(x=4\) (b) Lower: \(x=-1\)
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