Chapter 2.2, Problem 127E

To determine

ToSolve: the inequality $2x^2-x\ge 1$ ,sketch the solution on real number line and verify the solution using graphic utility.

Answer to Problem 127E

  $x=-\frac{1}{2}\text{and }x=1$

Explanation of Solution

Given:

Inequality is given as.

  $2x^2-x\ge 1$

Calculation:

Writing quadratic inequality in standard form.

  $\begin{array}{l}2x^2-x\ge 1\\ 2x^2-x-1\ge 0\end{array}$

Solving quadratic equation $2x^2-x-1=0$ .

  $\begin{array}{l}2x^2-x-1=0\\ 2x^2-\left(2-1\right)x-1=0\\ 2x^2-2x+x-1=0\\ 2x\left(x-1\right)+1\cdot \left(x-1\right)=0\\ \left(x-1\right)\left(2x+1\right)=0\\ x=1\text{ and }x=-\frac{1}{2}\end{array}$

  $x=1\text{ and }x=-\frac{1}{2}$ are the boundary points.

Since, our quadratic equation is greater than zero therefore we need intervals where it is positive.

Check for $x=0$ .

  $\begin{array}{l}2x^2-x-1\ge 0\\ 2{\left(0\right)}^2-0-1\ge 0\\ 0-0-1\ge 0\\ -1\ge 0\end{array}$

Since, it is negative therefor $x=0$ will be solution.

Check for $x=-1$ .

  $\begin{array}{l}2x^2-x-1\ge 0\\ 2{\left(-1\right)}^2-\left(-1\right)-1\ge 0\\ 2+1-1\ge 0\\ 2\ge 0\end{array}$

  $x=-1$ will be in the interval.

Check for $x=1$ .

  $\begin{array}{l}2x^2-x-1\ge 0\\ 2{\left(1\right)}^2-\left(1\right)-1\ge 0\\ 2-1-1\ge 0\\ 0\ge 0\end{array}$

  $x=1$ is also possible.

Therefore, from above check solution of inequality can be found as shown below.

  $(-\infty ,-\frac{1}{2}]\cup [1,\infty )$

It is plotted on real number line as.

  

Graph has been plotted using graphic utility as shown below.

  

It is clear that region right to the $x=-0.5$ and left to the $x=1$ are shaded therefore it is the solution of the inequality.

So, the answer is $x=-\frac{1}{2}\text{and }x=1$.