Chapter 2.2, Problem 130E

To determine

To solve:The given inequality on the real number line and verify the solution using graphing utility.

Answer to Problem 130E

The real numbers $x\in \left[-\infty ,-\frac{1}{2}\right]\cup \left[1,\infty \right]$ are the solutions of the inequality $2x^2-x\ge 1$ .

Explanation of Solution

Given data:

The given inequality is $2x^2-x\ge 1$ .

Method used:

Basic mathematical operations.

Calculations:

The given inequality is $2x^2-x\ge 1$

  $\Rightarrow 2x^2-x-1\ge 0$

  $\Rightarrow 2x^2-2x+x-1\ge 0$

  $\Rightarrow 2x\left(x-1\right)+1\left(x-1\right)\ge 0$

  $\Rightarrow \left(x-1\right)\left(2x+1\right)\ge 0$

Now, $\left(x-1\right)\left(2x+1\right)=0\Rightarrow x=1,x=-\frac{1}{2}$

Thus, the solutions of the inequality is $x<-\frac{1}{2},x<1$ and $x>-\frac{1}{2},x>1$ out of these solutions $x<-\frac{1}{2}$ and $x>1$ are acceptable as $\left(x-1\right)\left(2x+1\right)\ge 0$ for $x<-\frac{1}{2}$ and $x>1$ or the intervals $\left[-\infty ,-\frac{1}{2}\right]$ and $\left[1,\infty \right]$ .

The interval on the real line where $\left(x-1\right)\left(2x+1\right)\ge 0$ are the solutions of inequality shown in Figure-1 here.

  

The solutions of the inequality with graphing utility is shown in Figure-2 here.

  

From graph, we see that solutions of inequality $2x^2-x\ge 1$ are real numbers in intervals $\left[-\infty ,-\frac{1}{2}\right]$ and $\left[1,\infty \right]$ .

Thus, the solutions $\left[-\infty ,-\frac{1}{2}\right]$ and $\left[1,\infty \right]$ of given inequality stands verified.

Conclusion:

The real numbers $x\in \left[-\infty ,-\frac{1}{2}\right]\cup \left[1,\infty \right]$ are the solutions of the inequality $2x^2-x\ge 1$ .