Chapter 6.5, Problem 76SR

To determine

To Solve:

  $x^4+6x^2-27=0$

Answer to Problem 76SR

  $\sqrt{3}\text{,-}\sqrt{3}\text{,3}i\text{,-3}i$

Explanation of Solution

Given Information:

Given equation to solve:

  $x^4+6x^2-27=0$

Formula Used:

• ZERO PRODUCT PROPERTY:

If ab=0

Than , either one of a or b is zero.

That is , a = 0 or b = 0.

• 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 $x^4+6x^2-27=0$

  $\begin{array}{l}{x}^4+6x^2-27=0\\ $ \Rightarrow {(x^2)}^2+6x^2-27=\mathrm{0................}(1)$Let{\text{ y=x}}^2\\ (1)\Rightarrow y^2+6y-27=0\\ \Rightarrow y^2+9y-3y-27=0\\ \Rightarrow y(y+9)-3(y+9)=\mathrm{0.............}[\text{Greatest common factor]}\\ \Rightarrow (y+9)(y-3)=\mathrm{0.......................}[Factors]\\ \text{By the Zero Product Property:}\\ y+9=0\text{ or y-3=0}\\ \text{y=-9 or y=3}\\ {\text{Replacing y=x}}^2\\ \Rightarrow x^2=-9{\text{ or x}}^2=3\\ \Rightarrow x=\pm \sqrt{-1}\sqrt{9}\text{=}\pm 3i\text{ or x=}\pm \sqrt{3}\\ \text{So, the solutions are }\sqrt{3}\text{,-}\sqrt{3}\text{,3}i\text{,-3}i\end{array}$

Since , the index is 2 , which is even , so it will have 2 roots.