Question

If \(a^2+7 a<0\), then which one of the following could be the value of \(a\) ? (A) -3 (B) 0 (C) 1 (D) 2 (E) 3

Short answer

The value of \(a\) could be -3.

Step-by-step solution

Understand the inequality

The given inequality is \(a^2 + 7a < 0\). The goal is to determine which value of \(a\) satisfies this inequality.

Factor the inequality

We factor the inequality \(a^2 + 7a < 0\) into \(a(a + 7) < 0\). This tells us that the product of two terms, \(a\) and \(a + 7\), must be negative.

Determine the points where the inequality changes sign

The critical points, where \(a(a + 7) = 0\), occur at \(a = 0\) and \(a = -7\). These are the points where the expression changes sign.

Test intervals determined by critical points

The critical points divide the number line into intervals \((-\infty, -7)\), \((-7, 0)\), and \((0, \infty)\). Choose test points from each interval, such as \(a = -8\), \(a = -1\), and \(a = 1\). Substitute into the factored inequality to check where it is negative:- For \(a = -8\), \((-8)(-8 + 7) = 8 > 0\).- For \(a = -1\), \((-1)(-1 + 7) = -6 < 0\).- For \(a = 1\), \((1)(1 + 7) = 8 > 0\).

Evaluate the options for validity

Check the given options to see which fall into the interval \((-7, 0)\), where the inequality is valid:- (A) \(a = -3\) lies in \((-7, 0)\). Substitute to verify: \((-3)(-3 + 7) = -3 \times 4 = -12 < 0\). Therefore, \(a = -3\) satisfies the inequality.- (B) \(a = 0\), (C) \(a = 1\), (D) \(a = 2\), (E) \(a = 3\) do not satisfy \(a(a + 7) < 0\), as they either lie outside the interval or make the expression non-negative.

Key concepts

Inequality Problem Solving

Solving inequalities is similar to solving equations, but with a few key differences. In an inequality like the one given, \(a^2 + 7a < 0\), the expression does not equal a particular value, it simply gives a range of values that make the inequality true. The task is to find out for which values of \(a\) the expression remains below zero.

This involves several steps:

• Understanding the inequality format and analyzing its terms.
• Using algebraic manipulations, such as factoring, to simplify the inequality.
• Determining where the inequality changes from being true to false.

These steps will guide you to find out which values satisfy the inequality condition.

Factoring Inequalities

Factoring is a useful method in solving quadratics and related inequality problems. In our problem, the inequality \(a^2 + 7a < 0\) needs to be rewritten by factoring. Factoring helps to express the equation in a simpler form like \(a(a + 7) < 0\).

This factors the quadratic inequality into a product of two binomials. It's crucial because it allows determination of where the expression could change from positive to negative or vice versa as \(a\) changes.

By factoring, we create an easier path to finding where the inequality holds true, simplifying the process of finding valid solutions for \(a\).

Critical Points Analysis

Critical points are values of \(a\) where the expression equals zero. For the factorized inequality \(a(a + 7) < 0\), the critical points occur at \(a = 0\) and \(a = -7\). These points are important because they are where the inequality changes sign.

Knowing these points allows us to divide the number line into distinct intervals:

• \((-\infty, -7)\)
• \((-7, 0)\)
• \((0, \infty)\)

In these intervals, the sign of the product \(a(a + 7)\) will be constant. We can then test where the product is negative, thus solving the inequality.

Interval Testing in Inequalities

Interval testing involves choosing representative values from the intervals created by the critical points \(a = 0\) and \(a = -7\). By substituting these values into the inequality \(a(a + 7) < 0\), we determine where the inequality holds.

For each interval:

• In \((-\infty, -7)\), pick \(a = -8\). The product \((-8)(-1)\) is positive.
• In \((-7, 0)\), pick \(a = -1\). The product \((-1)(6)\) is negative.
• In \((0, \infty)\), pick \(a = 1\). The product \((1)(8)\) is positive.

These results indicate the solution set is \((-7, 0)\). Within this interval, the inequality \(a(a+7) < 0\) is satisfied, thus pinpointing viable solutions to our problem.