Chapter 4.4, Problem 45AYU

To determine

To calculate:

The solution of the following inequality equation:

  $\frac{x^2\left(3+x\right)\left(x+4\right)}{\left(x+5\right)\left(x-1\right)}\ge 0$

Answer to Problem 45AYU

The solution of the inequality equation $\frac{x^2\left(3+x\right)\left(x+4\right)}{\left(x+5\right)\left(x-1\right)}\ge 0$ is $x=0,-3,-4$ .

Explanation of Solution

Given information:

  $\frac{x^2\left(3+x\right)\left(x+4\right)}{\left(x+5\right)\left(x-1\right)}\ge 0$

Calculation:

The given equation is:

  $\begin{array}{l}\frac{x^2\left(3+x\right)\left(x+4\right)}{\left(x+5\right)\left(x-1\right)}\ge 0\\ \frac{x^2\left(3+x\right)\left(x+4\right)}{\left(x+5\right)\left(x-1\right)}=0\\ x^2\left(3+x\right)\left(x+4\right)=0OR\text{}\left(x+5\right)\left(x-1\right)=0\\ x^2=0OR\text{}3+x=0ORx+4=0OR\text{}x+5=0ORx-1=0\\ x=0x=-3x=-4x=-5x=1\end{array}$

Need the inequality value which is greater than or equal to zero, from the above solution $\begin{array}{l}x\ge 0,1\\ x<0,-5,-4,-3<0\end{array}$

Hence, the inequality is solved for $x=0,1$ .