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Chapter 2: Problem 24

Solve the inequality. Then graph the solution set. $$-2 x^{2}+6 x \leq-15$$

Short Answer

Expert verified
The solution to the inequality is \(3 - \sqrt{6} \leq x \leq 3 + \sqrt{6}\). The solution set, when graphed on a number line, shows shading between \(3-\sqrt{6}\) to \(3+\sqrt{6}\) inclusive.

Step by step solution

01

Rewrite the inequality in standard form

To accomplish this, the inequality \(-2 x^{2}+6 x \leq -15\) can be rewritten as \(-2x^2 + 6x + 15 \leq 0\). It is important to ensure that the inequality looks like a standard quadratic equation.
02

Solve the corresponding quadratic equation

Let's find the roots of the equation \(-2x^2 + 6x + 15 = 0\). Use the quadratic formula to solve for x, which is \(x = [-b 卤 sqrt(b^2 -4ac)] / (2a)\). Therefore, the two roots \(x_1\) and \(x_2\) are \(x_1= 3 + \sqrt{6}\) and \(x_2=3-\sqrt{6}\).
03

Perform the interval test

After finding the roots, divide the number line into three intervals using these roots. For the given inequality, these intervals are \((-鈭�, 3 - \sqrt{6}]\), \([3 - \sqrt{6}, 3 + \sqrt{6}]\) and \([3 + \sqrt{6}, 鈭�)\). Select a test point from each interval and substitute it into the inequality. If the inequality is satisfied, then that interval is part of the solution set.
04

Graph the solution set

On the number line, indicate the solution set for the inequality. The roots of the quadratic equation represent the boundary points. If the inequality is 'less than or equal to', then you include the boundary points in the solution set by using a closed circle. If not, you represent the boundary point with an open circle. As for this inequality, since it is 'less than or equal to', we will use a closed circle. Shade the intervals which satisfied the inequality during the interval test. For this inequality, the number line will be shaded within the interval from \(3 - \sqrt{6}\) to \(3 + \sqrt{6}\), including the end points.

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Most popular questions from this chapter

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