Problem 36 - Intermediate Algebra

Question

Solve each inequality. $$ 2 x^{2}+6 x<0 $$

Step-by-Step Solution

Answer

The solution is \( x \in (-3, 0) \).

Step 1: Factor the inequality

Start by factoring the left side of the inequality. The expression can be factored as follows: \( 2x^2 + 6x = 2x(x + 3) \). The inequality now looks like \( 2x(x + 3) < 0 \).

Step 2: Determine the critical points

Find the values of \(x\) that make each factor equal to zero. This will help us determine the intervals to test.1. Solve for \(2x = 0\): \[ x = 0 \] 2. Solve for \(x + 3 = 0\): \[ x = -3 \] The critical points are \( x = -3 \) and \( x = 0 \).

Step 3: Test intervals using critical points

Now test values within the intervals determined by the critical points: 1. Interval \((-\infty, -3)\): Choose \(x = -4\). \(2(-4)((-4) + 3) = 2(-4)(-1) = 8 > 0\)2. Interval \((-3, 0)\): Choose \(x = -1\). \(2(-1)((-1) + 3) = 2(-1)(2) = -4 < 0\)3. Interval \((0, \infty)\): Choose \(x = 1\). \(2(1)((1) + 3) = 2(1)(4) = 8 > 0\)

Step 4: Analyze the results

From testing the intervals, the inequality \(2x(x + 3) < 0\) is satisfied in the interval \((-3, 0)\). This means that for any \(x\) in this range, the inequality will hold true.

Key Concepts

Factoring Polynomials

When faced with a polynomial expression, especially in an inequality, the first step is often to simplify it by factoring. To factor a polynomial, we break down the expression into smaller "factors" that multiply together to give the original polynomial.
In the example, we have the polynomial expression \(2x^2 + 6x\). The key to factoring is to identify common factors among the terms.

• Look for the greatest common factor (GCF) of all the terms in the expression.
• In \(2x^2 + 6x\), both terms share a factor of \(2x\).
• By factoring out \(2x\), we rewrite the expression as \(2x(x + 3)\).

This process turns our expression into a product of factors, which is much easier to work with in an inequality context.

Quadratic Inequalities

Quadratic inequalities, like \(2x(x + 3) < 0\), involve expressions with a degree of two. They are similar to quadratic equations but instead use inequalities such as \(<, >, \leq, \geq\) instead of an equals sign.
The goal is to find the range of values for \(x\) that make the inequality true.

• To deal with these, we first write the polynomial in a factored form.
• This makes it easier to identify critical points, which are where each factor equals zero.

Unlike a simple quadratic equation, the inequality involves testing regions or intervals within the critical points, as values in these intervals might either make the inequality true or false. In our example, the factored form \(2x(x+3) < 0\) introduces critical points at \(x = 0\) and \(x = -3\).

Interval Testing

Once the critical points are found, we use these points to divide the number line into intervals. Interval testing is the method of checking which intervals satisfy the inequality.
For the inequality \(2x(x + 3) < 0\), we have critical points at \(x = -3\) and \(x = 0\). These points split the number line into three intervals:

• \(( -\infty, -3 )\)
• \(( -3, 0 )\)
• \(( 0, \infty )\)

We then select a test point from each interval and substitute it back into the factored inequality. For instance:

• In \(( -\infty, -3 )\), try \(x = -4\): results in positive, \(8 > 0\).
• In \(( -3, 0 )\), try \(x = -1\): results in negative, \(-4 < 0\).
• In \(( 0, \infty )\), try \(x = 1\): results in positive, \(8 > 0\).

Only the interval \((-3, 0)\) makes the inequality true. Thus, the solution of the inequality occurs within this interval, confirming where \(2x(x + 3)\) is less than zero.