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Chapter 3: Problem 17

Solve each polynomial inequality in Exercises \(1-42\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ 5 x \leq 2-3 x^{2} $$

Short Answer

Expert verified
The solution set is \(-\infty, -1\] \cup \[2/3, +\infty\).

Step by step solution

01

Rearranging the inequality

First, rearrange the given inequality into a standard quadratic format. This can be done by adding \(3x^2\) and subtracting \(5x\) on both sides resulting in: \n$$3x^2 + 5x - 2 \geq 0.$$
02

Find the roots

Next, apply the quadratic formula to find the roots. The quadratic formula is named as such: \(-b\pm\sqrt{b^2 - 4ac} \over 2a\). For this inequality, \(a = 3\), \(b = 5\) and \(c = -2\). Now, calculate the roots with these values and this results in the roots: \(x_1 = -1\) and \(x_2 = 2/3\).
03

Determine the intervals

The inequality is divided into three intervals by the roots, \(-\infty, -1\), \(-1, 2/3\), and \(2/3, +\infty\). Pick a test point in each interval and substitute it into the inequality. If the inequality holds, then that interval is part of the solution set. In this case, the test points could be -2, 0 and 1. Substituting shows that the intervals \(-\infty, -1\] and \[2/3, +\infty\) are part of the solution.
04

Solution in interval notation

Lastly, write the solution in interval notation. This gives the result \(-\infty, -1\] \cup \[2/3, +\infty\).

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