Chapter 5.8, Problem 22WE

To determine

To find the open sentence of $\frac{z^2+1}{6}\ge \frac{z+2}{3}$

Answer to Problem 22WE

The value of open sentence is $z:z\le -1orz\ge 3$

Explanation of Solution

Given open sentence is $\frac{z^2+1}{6}\ge \frac{z+2}{3}$

Value of open sentence means a sentence is it true and depend upon the variable values .

  $\frac{z^2+1}{6}\ge \frac{z+2}{3}$

Multiply both the sides by 6, the LCM of 6 and 3

  $6\left(\frac{z^2+1}{6}\right)\ge 6\left(\frac{z+2}{3}\right)$

Apply distributive property and simplify $\left[a(b\pm c)=a\ast b\pm a*c\right]$

  $z^2+1\ge 2z+4$

  $\begin{array}{l}\text{Subtract}2z+4\text{from}\text{both}\text{sides}\\ z^2+1-2z-4\ge 0\\ z^2-2z-3\ge 0\end{array}$

Use factorization and simplify

  $\begin{array}{c}{z}^2-2z-3\ge 0\\ z^2-(3-1)z-3\ge 0\\ z^2-3z+z-3\ge 0\\ z(z-3)+1(z-3)\ge 0\\ (z-3)(z+1)\ge 0\end{array}$

Now, check the variable value satisfied the sentence with zero product property with inequality .

  $\begin{array}{c}(z-3)\ge 0\\ z\ge 3\\ \text{Similarly}\\ (z+1)\ge 0\\ z\le -1\end{array}$

Both factor values satisfied in the given condition, therefore, need to summarize the table:

	$z=3$	$z\succ 3$	$z=-1$	$z>-1$
$z-3$	0	+	-	-
$z+1$	+	+	0	-
$(z-3)(z-1)$	0	+	0	-

Hence, the open sentence variable values $z:z\le -1orz\ge 3$ satisfy the required condition $\le 0$ where both factor have the same sign.