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Chapter 2: Problem 19

Solve the inequality. Then graph the solution set. $$x^{2}+x<6$$

Short Answer

Expert verified
The solution set to the inequality is \((-鈭�,-3) \cup (2,鈭�)\)

Step by step solution

01

Rearrange the Inequality

Subtract 6 from both sides of the inequality to bring it to a standard quadratic form as so: \(x^{2}+x-6<0\)
02

Factor the Quadratic Expression

This quadratic factors into \( (x-2)(x+3)<0\)
03

Find the Roots or Critical Values

Set \( (x-2)(x+3)=0 \), solving for x gives the roots as \( x=2 \) and \( x=-3 \)
04

Create Intervals

The roots divide the number line into three intervals, \((-鈭�,-3)\), \((-3,2)\) and \((2,鈭�)\).
05

Test Values in Each Interval

Choose test points from each interval and substitute these test points into the inequality. If the inequality holds true then that interval is part of the solution set.
06

Draw the Graph

Show the solutions on the number line. Open circles are used at x = -3 and x = 2 because the inequality is 'less than' not 'less than or equal to'.

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Most popular questions from this chapter

Think About It \(\quad\) A cubic polynomial function \(f\) has real zeros \(-2, \frac{1}{2},\) and \(3,\) and its leading coefficient is negative. Write an equation for \(f\) and sketch its graph. How many different polynomial functions are possible for \(f ?\)

Find all the zeros of the function. When there is an extended list of possible rational zeros, use a graphing utility to graph the function in order to disregard any of the possible rational zeros that are obviously not zeros of the function. $$f(x)=9 x^{3}-15 x^{2}+11 x-5$$

Find the rational zeros of the polynomial function. $$f(x)=x^{3}-\frac{1}{4} x^{2}-x+\frac{1}{4}=\frac{1}{4}\left(4 x^{3}-x^{2}-4 x+1\right)$$

Use Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of the function. $$f(x)=3 x^{3}+2 x^{2}+x+3$$

Write the polynomial as the product of linear factors and list all the zeros of the function. $$f(x)=x^{2}-x+56$$
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