Chapter 4, Problem 33SGA

To determine

Find the possible rational roots of the equation.

Answer to Problem 33SGA

  $\pm 1,\pm 5$ ; $\pm 1$

Explanation of Solution

Given information:

  $x^4+4x^2-5=0$

Calculation:

Here, we have:

  $x^4+4x^2-5=0$

Now, according to the rational root theorem, if $\frac{p}{q}$ is a root of the equation, then $p$ is a factor of $5andq$ is a factor of $1$ .

Thus,

The possible values of p:

  $\pm 1,\pm 5$

The possible values of $q$ :

  $\pm 1$

The possible rational roots $\frac{p}{q}$ :

  $\pm 1,\pm 5$

Now, we know that all the possible rational roots fall in the domain $-5\le x\le 5$ .

Thus, set the $x-axis$ viewing window at $[-6,6]$ .

Now, graph the function:

  

Now, by synthetic division method to check $-1$ :

  $\begin{array}{llllll}-1\hfill & 1\hfill & 0\hfill & 4\hfill & 0\hfill & -5\hfill \\ \hfill & \hfill & -1\hfill & 1\hfill & -5\hfill & 5\hfill \\ \hfill & 1\hfill & -1\hfill & 5\hfill & -5\hfill & 0\hfill \end{array}$

Now, by synthetic division method to check $1$ :

  $\begin{array}{lllll}1\hfill & 1\hfill & -1\hfill & 5\hfill & -5\hfill \\ \hfill & \hfill & 1\hfill & 0\hfill & 5\hfill \\ \hfill & 1\hfill & 0\hfill & 5\hfill & 0\hfill \end{array}$

Therefore, $x^4+4x^2-5=\left(x+1\right)\left(x-1\right)\left(x^2+5\right)$ .

Now, solving for $x^2+5$ :

  $\begin{array}{l}{x}^2+5\\ =x^2-{\left(\sqrt{5}i\right)}^2\\ =\left(x+\sqrt{5}i\right)\left(x-\sqrt{5}i\right)\end{array}$

The roots of the equation $x^4+4x^2-5=0$ are:

  $-\sqrt{5}i,\sqrt{5}i,-1and1$

Hence, the rational roots of the equation are $\pm 1,\pm 5$ ; $\pm 1$

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