Chapter 7.6, Problem 67HP

(a)

To determine

Explain why the given expression is not a real number.

Explanation of Solution

Given:

The expression is :

  $\sqrt[4]{{\left(-16\right)}^3}$

Calculation:

Given:

  $\begin{array}{l}\sqrt[4]{{\left(-16\right)}^3}\\ =\sqrt[4]{-{\left(2^4\right)}^3}\mathrm{..................}[{\left(-1\right)}^3=-1]\\ =\left(2^3\right)\sqrt[4]{-1}\\ =8\sqrt[4]{-1}\end{array}$

Since the index is 4 , which is even , so the radicand cannot be negative.

Here the radicand is negative.

Hence the number is imaginary and not real .

(b)

To determine

Find n such that $n\sqrt[4]{{\left(-16\right)}^3}$ is a real number.

Answer to Problem 67HP

  $n=\sqrt[4]{-1}$

Explanation of Solution

Given:

The expression is :

  $n\sqrt[4]{{\left(-16\right)}^3}$

Calculation:

Given:

  $\begin{array}{l}n\sqrt[4]{{\left(-16\right)}^3}\\ =n\sqrt[4]{-{\left(2^4\right)}^3}\mathrm{..................}[{\left(-1\right)}^3=-1]\\ =n\left(2^3\right)\sqrt[4]{-1}\\ =8n\sqrt[4]{-1}\end{array}$

Since the index is 4 , which is even , so the radicand cannot be negative.

So, to make it a real number, we need to eliminate the negative sign under the radical .

Let $n=\sqrt[4]{-1}$

Check:

  $\begin{array}{l}n\sqrt[4]{{\left(-16\right)}^3}\\ =8n\sqrt[4]{-1}\\ =8\cdot \sqrt[4]{-1}\cdot \sqrt[4]{-1}\mathrm{................}[n=\sqrt[4]{-1}]\\ =8\sqrt[4]{\left(-1\right)\cdot \left(-1\right)}\mathrm{...............}[\text{Product property of radicals]}\\ =8\sqrt[4]{1}\\ =8\end{array}$

And 8 is a real number.

So, $n=\sqrt[4]{-1}$