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Chapter 2: Problem 13

Solve each polynomial inequality and graph the solution set on a real number line. Express each solution set in interval notation. $$2 x^{2}+x<15$$

Short Answer

Expert verified
The solution to the inequality is \((-3, \frac{5}{2})\).

Step by step solution

01

Simplify the inequality

First, re-arrange the given equation to one side: \(2x^2 + x - 15 < 0\). This will then enable us to factorize the quadratic expression.
02

Factorize the expression

Now, factorize the expression on the left-hand side. That is \(2x^2 + x - 15 = (2x -5)(x+3)\). Always remember the rule of Factoring. For any real numbers a, b, and nonzero number c, \(acx^2 + (ad + bc)x + bd = (cx + d)(ax + b)\). This results in the factored inequality \((2x - 5)(x + 3) < 0\).
03

Identify critical points

Critical points are the x-values that make the expression zero. Here, they are \(\frac{5}{2}\) and \(-3\) gotten by equaling to zero \((2x - 5)\) and \((x + 3)\) respectively. These two points divide the number line into three intervals.
04

Test the intervals

For each interval, choose a test point and substitute back into the inequality to see whether the statement is true or false. The intervals to be tested are \((-\infty, -3)\), \((-3, \frac{5}{2})\), \((\frac{5}{2}, \infty)\). Testing these intervals, we realise that the inequality holds true for \(-3 < x < \frac{5}{2}\).
05

Write the solution in interval notation and graph

Now that we know the values of x that satisfy the inequality, we can write this in interval notation as \((-3, \frac{5}{2})\). This means all values of x between -3 and \(\frac{5}{2}\), exclusive. We can also represent this graphically on a number line with open circles to represent the endpoints, and an arrow pointing right to indicate the interval between -3 and \(\frac{5}{2}\).

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