How many integer values of n satisfy the inequality |n^2 - 9| \leq 2n + 6? — GMAT

How many integer values of $n$ satisfy the inequality $|n^2 - 9| \leq 2n + 6$ ?

Answer: 6

Answer

There are 6 integer values of

n

that satisfy the given inequality.

Because

|n^2 - 9| \geq 0

for all real numbers,

2n + 6

must be non-negative, requiring

n \geq -3

. At

n = -3

, the inequality reduces to

0 \leq 0

, which is valid. For

n > -3

, the term

n + 3

is positive, so factoring

|n^2 - 9|

into

|n - 3|(n + 3)

and dividing by

n + 3

simplifies the inequality to

|n - 3| \leq 2

. This yields the integer range

1 \leq n \leq 5

, containing 5 integers. Including

n = -3

gives a total of 6 integer solutions.

Step-by-Step Solution

Determine the domain constraint from the non-negativity of the absolute value expression.

2n + 6 \geq 0 \implies n \geq -3

The absolute value expression

|n^2 - 9|

is non-negative for all real

n

, so the right-hand side must also be non-negative.

Evaluate the boundary case

n = -3

|(-3)^2 - 9| = 0

and

2(-3) + 6 = 0 \implies 0 \leq 0

(True)

When

n = -3

, both sides equal 0, making

n = -3

a valid integer solution.

Factor the quadratic inside the absolute value for

n > -3

and simplify.

|n - 3||n + 3| \leq 2(n + 3) \implies |n - 3| \leq 2

Since

n > -3

, the term

n + 3

is strictly positive, allowing division of both sides by

n + 3

without reversing the inequality.

Solve the linear absolute value inequality

|n - 3| \leq 2

-2 \leq n - 3 \leq 2 \implies 1 \leq n \leq 5

Removing the absolute value creates a compound inequality bounded between

-2

and

2

List all valid integer solutions and count them.

n \in \{-3, 1, 2, 3, 4, 5\} \implies 6

integer solutions

Combining the boundary root

n = -3

with the five consecutive integers from

1

5

gives 6 integer solutions.

Key Concept

Absolute Value Inequalities with Variable RHS and Boundary Factors

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