Chapter 6.5, Problem 77SR

To determine

To Solve:

  $x^3+64=0$

Answer to Problem 77SR

  $2+2i\sqrt{3},2-2i\sqrt{3},-4$

Explanation of Solution

Given Information:

Given equation to solve:

  $x^3+64=0$

Formula Used:

• ZERO PRODUCT PROPERTY:

If ab=0

Than , either one of a or b is zero.

That is , a = 0 or b = 0.

• $a^3+b^3=(a+b)(a^2-ab+b^2)$
• Definition of nth root:
• Words : For any real numbers a and b , and any positive integer n , if $a^n=b$ then a is the nth root of b . That is $a=\sqrt[n]{b}$

$\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$

• The non-negative root is called the Principal Root.
• Suppose n is an integer greater than 1, and a is a real number.

a	n is even	n is odd
a > 0	1 unique positive and 1 unique negative real root : $\pm \sqrt[n]{a}$ ; positive root is principal root.	1 unique positive and 0 negative real root : $\sqrt[n]{a}$
a < 0	0 real roots	0 positive and 1 negative real root : $-\sqrt[n]{a}$
a=0	1 real root : $\sqrt[n]{0}=0$	1 real root : $\sqrt[n]{0}=0$

• Pure Imaginary Numbers :
• Square roots of negative real numbers: for any positive real number b , $\sqrt{-b^2}=\sqrt{b^2}\sqrt{-1}$ , or $$ bi where

$\begin{array}{l}\sqrt{-1}=i\\ i=iota\end{array}$

Calculation:

• Given $x^3+64=0$

  $\begin{array}{l}{x}^3+64=0\\ \Rightarrow x^3+4^3=0\\ \Rightarrow (x+4)(x^2-4x+4^2)=\mathrm{0...........}[a^3+b^3=(a+b)(a^2-ab+b^2)]\\ \text{Using the zero product property:}\\ \text{(x+4)=0 }\\ \Rightarrow x=-4\\ \text{or }\\ {\text{(x}}^2-4x+16)=0\\ {\text{Comparing it with ax}}^2+bx+c=0\\ a=1,b=-4,c=16\\ Roots=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-(-4)\pm \sqrt{16-4(1)(16)}}{2(1)}\\ =\frac{4\pm \sqrt{-48}}{2}\\ =\frac{4\pm 4i\sqrt{3}}{2}=2\pm 2i\sqrt{3}\end{array}$

Since , the degree of equation is 3 , so it will have 3 roots.

The solution is $2+2i\sqrt{3},2-2i\sqrt{3},-4$