Chapter 8, Problem 7CA

a.

To determine

Find the solution of the given equation and check whether the results are a real numbers, imaginary numbers or pure imaginary numbers.

Answer to Problem 7CA

The solution of the equation is a pure imaginary number.

Explanation of Solution

Given:

The given equation is $x+\sqrt{-16}=0$ .

Calculation:

  $x+\sqrt{-16}=0$

Apply radical property $\sqrt{-a}=\sqrt{-1}\sqrt{-a}$ and $i=\sqrt{-1}$ .

  $\begin{array}{l}\Rightarrow x+\sqrt{-1}\cdot \sqrt{16}=0\\ \Rightarrow x+i\cdot 4=0\\ \Rightarrow x+4i=0\end{array}$

Subtract $4i$ from both sides.

  $\begin{array}{l}\Rightarrow x+4i-4i=0-4i\\ \Rightarrow x=-4i\end{array}$

Hence the solution of the equation is a pure imaginary number.

b.

To determine

Find the solution of the given equation and check whether the results are a real numbers, imaginary numbers or pure imaginary numbers.

Answer to Problem 7CA

The solution of the equation is an imaginary number.

Explanation of Solution

Given:

The given equation is $\left(11-2i\right)-\left(-3i+6\right)=8+x$ .

Calculation:

  $\left(11-2i\right)-\left(-3i+6\right)=8+x$

Remove parentheses.

  $\begin{array}{l}\Rightarrow 11-2i+3i-6=8+x\\ \Rightarrow 5+i=8+x\end{array}$

Subtract $8$ from both sides.

  $\begin{array}{l}\Rightarrow 5+i-8=8+x-8\\ \Rightarrow -3+i=x\\ \Rightarrow x=-3+i\end{array}$

Hence the solution of the equation is an imaginary number.

c.

To determine

Find the solution of the given equation and check whether the results are a real numbers, imaginary numbers or pure imaginary numbers.

Answer to Problem 7CA

The solutions of the equation are a pure imaginary numbers.

Explanation of Solution

Given:

The given equation is $3x^2-14=-20$ .

Calculation:

  $3x^2-14=-20$

Add $14$ to both sides.

  $\begin{array}{l}\Rightarrow 3x^2-14+14=-20+14\\ \Rightarrow 3x^2=-6\end{array}$

Divide both sides by $3$ .

  $\begin{array}{l}\Rightarrow \frac{3x^2}{3}=-\frac{6}{3}\\ \Rightarrow x^2=-2\\ \Rightarrow x=\pm \sqrt{-2}\\ i=\sqrt{-1}\\ \Rightarrow x=\pm \sqrt{-1}\cdot \sqrt{2}\\ \Rightarrow x=\pm i\cdot \sqrt{2}\\ \Rightarrow x=\pm \sqrt{2}i\end{array}$

Hence the solutions of the equation are a pure imaginary numbers.

d.

To determine

Find the solution of the given equation and check whether the results are a real numbers, imaginary numbers or pure imaginary numbers.

Answer to Problem 7CA

The solutions of the equation are an imaginary numbers.

Explanation of Solution

Given:

The given equation is $x^2+2x=-3$ .

Calculation:

  $x^2+2x=-3$

Add $3$ to both sides.

  $\begin{array}{l}\Rightarrow x^2+2x+3=-3+3\\ \Rightarrow x^2+2x+3=0\end{array}$

Use quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

In the given equation, $a=1,b=2,c=3$ .

  $\begin{array}{l}x=\frac{-2\pm \sqrt{2^2-4\cdot 1\cdot 3}}{2\cdot 1}\\ =\frac{-2\pm \sqrt{4-12}}{2}\\ =\frac{-2\pm \sqrt{-8}}{2}\\ =\frac{-2\pm 2\sqrt{-2}}{2}\\ =-1\pm \sqrt{-2}\\ i=\sqrt{-1}\\ =-1\pm \sqrt{-1}\cdot \sqrt{2}\\ =-1\pm \sqrt{2}i\end{array}$

Hence the solutions of the equation are an imaginary numbers.

e.

To determine

Find the solution of the given equation and check whether the results are a real numbers, imaginary numbers or pure imaginary numbers.

Answer to Problem 7CA

The solutions of the equation are real numbers.

Explanation of Solution

Given:

The given equation is $x^2=16$ .

Calculation:

  $x^2=16$

  $\begin{array}{l}\Rightarrow x=\pm \sqrt{16}\\ \Rightarrow x=\pm 4\end{array}$

Hence the solutions of the equation are real numbers.

f.

To determine

Find the solution of the given equation and check whether the results are a real numbers, imaginary numbers or pure imaginary numbers.

Answer to Problem 7CA

The solutions of the equation are real numbers.

Explanation of Solution

Given:

The given equation is $x^2-5x-8=0$ .

Calculation:

  $x^2-5x-8=0$

Use quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

In the given equation, $a=1,b=-5,c=-8$ .

  $\begin{array}{l}x=\frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-4\cdot 1\cdot \left(-8\right)}}{2\cdot 1}\\ =\frac{5\pm \sqrt{25+32}}{2}\\ =\frac{5\pm \sqrt{57}}{2}\end{array}$

Hence the solutions of the equation are real numbers.