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Magazine Chapter 1: Problem 43
Nonlinear Inequalities Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$2 x^{2}+x \geq 1$$
Short Answer
Expert verified
The solution is \([-1, -1] \cup [\frac{1}{2}, \infty)\).
Step by step solution
01
Rearrange the Inequality
First, rearrange the inequality to set it to 0 on one side. Start with the given inequality: \(2x^2 + x \geq 1\). Subtract 1 from both sides to get \(2x^2 + x - 1 \geq 0\).
02
Find the Roots of the Corresponding Equation
To solve the inequality \(2x^2 + x - 1 \geq 0\), first find the roots of the corresponding equation \(2x^2 + x - 1 = 0\). Use either factoring, completing the square, or the quadratic formula to solve for \(x\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), and \(c = -1\), calculate the roots: \(x = \frac{-1 \pm \sqrt{(1)^2 - 4(2)(-1)}}{4}\). Simplify to find \(x = \frac{-1 \pm \sqrt{1 + 8}}{4}\), or \(x = \frac{-1 \pm 3}{4}\). Thus, the roots are \(x = \frac{1}{2}\) and \(x = -1\).
03
Determine Test Intervals
The roots \(x = \frac{1}{2}\) and \(x = -1\) divide the number line into the intervals \((-\infty, -1)\), \((-1, \frac{1}{2})\), and \((\frac{1}{2}, \infty)\). Use these intervals to test values and determine where the inequality holds true.
04
Test Each Interval
Select test points from each interval. For \((-\infty, -1)\), use \(x = -2\): \(2(-2)^2 + -2 - 1 = 8 - 2 - 1 = 5\). Since 5 is positive, this interval is part of the solution set. For \((-1, \frac{1}{2})\), use \(x = 0\): \(2(0)^2 + 0 - 1 = -1\). Since -1 is negative, this interval is not part of the solution. For \((\frac{1}{2}, \infty)\), use \(x = 1\): \(2(1)^2 + 1 - 1 = 2\). Since 2 is positive, this interval is part of the solution.
05
Write the Solution in Interval Notation
Combine the intervals where the inequality is true. The solution includes \((-\infty, -1)\) and \((\frac{1}{2}, \infty)\). The inequality includes the endpoints where the expression equals zero, which are \(x = -1\) and \(x = \frac{1}{2}\). Therefore, the complete solution in interval notation is \([-1, -1] \cup [\frac{1}{2}, \infty)\).
06
Graph the Solution Set
To graph the solution set, plot the points \(x = -1\) and \(x = \frac{1}{2}\) on a number line. Use a closed dot at \(x = -1\) to indicate that it is included in the solution set, and an open dot at \(x = \frac{1}{2}\) (in corrected solution, closed dot is used) for the infinity end, with shading to the left of \(x = -1\) and shading to the right of \(x = \frac{1}{2}\) toward positive infinity.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a tool we use to find the roots of a quadratic equation, which is any equation in the form: \( ax^2 + bx + c = 0 \).
Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula allows you to directly calculate the values of \( x \) where the quadratic expression equals zero.
You simply need to plug in the values of \( a \), \( b \), and \( c \) from your equation.
Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula allows you to directly calculate the values of \( x \) where the quadratic expression equals zero.
You simply need to plug in the values of \( a \), \( b \), and \( c \) from your equation.
- In our example, for \( 2x^2 + x - 1 = 0 \), \( a = 2 \), \( b = 1 \), and \( c = -1 \).
- By substituting these values into the formula, we calculated the roots to be \( x = -1 \) and \( x = \frac{1}{2} \).
Interval Notation
Interval notation is a concise way of describing a set of numbers鈥攖he solutions to our inequality鈥攖hat encompass a certain range on the number line.
This method employs round brackets \( () \) when excluding endpoints and square brackets \( [] \) to include them.
For nonlinear inequalities like \( 2x^2 + x - 1 \geq 0 \), finding where the expression is satisfied involves:
This method employs round brackets \( () \) when excluding endpoints and square brackets \( [] \) to include them.
For nonlinear inequalities like \( 2x^2 + x - 1 \geq 0 \), finding where the expression is satisfied involves:
- Locating the roots from the quadratic equation.
- Testing intervals between these roots to see where the inequality holds true.
- For the equation \( 2x^2 + x - 1 = 0 \) with roots \( -1 \) and \( \frac{1}{2} \), we analyzed intervals around these points.
- We found that \(( -\infty, -1 ]\) and \([ \frac{1}{2}, \infty )\) are part of the solution.
Graphing Solutions
Graphing solutions of inequalities on a number line gives us a visual representation of the solution set.
For \( 2x^2 + x - 1 \geq 0 \), which is a quadratic inequality, we proceed as follows:
For \( 2x^2 + x - 1 \geq 0 \), which is a quadratic inequality, we proceed as follows:
- Mark the calculated roots \( x = -1 \) and \( x = \frac{1}{2} \) on the number line with closed dots since these points satisfy the equality part of our inequality (\( \geq \)).
- Shade the regions where the inequality holds true: to the left of \(-1\) and to the right of \(\frac{1}{2}\). These regions represent all the \( x \) values that satisfy the inequality.
- The section from \(-\infty\) to \(-1\), shaded and with a closed dot, indicates inclusion of \(-1\). This shows all values are solutions.
- Similarly, from \(\frac{1}{2}\) to \(\infty\), with shading and a closed dot indicates the inclusion of \(\frac{1}{2}\), showing continuity of solutions beyond this point.
