Chapter 0.2, Problem 2QR

To determine

To find: Solve the inequality for $x$ .

Answer to Problem 2QR

The solution of given inequality is $x<0$ or $x>2$ .

Explanation of Solution

Given information:

The given inequality is $x\left(x-2\right)>0$ .

Calculation:

Rewrite

  $\begin{array}{l}x\left(x-2\right)>0\\ x^2-2x>0\end{array}$

Find the critical point of an inequality.

  $\begin{array}{l}{x}^2-2x=0\\ x(x-2)=0\\ x=0,x=2\end{array}$

There are three possibilities $x<0,0<x<2,x>2$ .

Now it is required to check interval between the critical points.

if $x<0$

let’s take any point which is less than 0 , say -1.

Substitute this value in the inequality:

  $\begin{array}{l}-1(-1-2)>0\\ -1\cdot \left(-3\right)>0\\ 3>0\end{array}$

This is true.Hence it satisfies the given inequality.

If $0<x<2$

Let take the point 1.

Substitute this value in the given inequality:

  $\begin{array}{l}1(1-2)>0\\ -1>0\end{array}$

This is false hence the case $0<x<2$ will be declined.

Now if $x>2$ . Take any point which is greater than 2 let it be 3.

Substitute this value in the given inequality:

  $\begin{array}{l}3(3-2)>0\\ 3>0\end{array}$

This is true hence this condition satisfies the inequality.

Therefore the solution for $x$ will be $x<0$ or $x>2$ .