If n is an integer such that |2n + 1| \le |n - 4|, how many distinct integer values of n s… — GMAT

If $n$ is an integer such that $|2n + 1| \le |n - 4|$ , how many distinct integer values of $n$ satisfy the inequality?

There are 7 distinct integer values of n that satisfy the inequality.

Squaring both non-negative sides of the absolute value inequality

|2n + 1| \le |n - 4|

eliminates the absolute value bars without requiring multiple case splits. Expanding

(2n + 1)^2 \le (n - 4)^2

yields

4n^2 + 4n + 1 \le n^2 - 8n + 16

, which simplifies to

3n^2 + 12n - 15 \le 0

. Dividing the inequality by 3 gives

n^2 + 4n - 5 \le 0

, which factors as

(n + 5)(n - 1) \le 0

. The inequality is satisfied for all values in the closed interval

[-5, 1]

. Listing the integers in this range yields

-5, -4, -3, -2, -1, 0, 1

, which comprises exactly 7 integers.

Square both sides of the inequality since absolute values are non-negative.

(2n + 1)^2 \le (n - 4)^2

Since

|A| \le |B|

is equivalent to

A^2 \le B^2

for all real numbers.

Expand both algebraic expressions and move all terms to one side.

4n^2 + 4n + 1 \le n^2 - 8n + 16 \implies 3n^2 + 12n - 15 \le 0

Standard quadratic inequality form requires comparing the quadratic polynomial to zero.

Simplify by dividing by 3 and factor the quadratic expression.

n^2 + 4n - 5 \le 0 \implies (n + 5)(n - 1) \le 0

Dividing by a positive constant preserves the inequality sign and allows factoring into linear binomials.

Determine the solution set for the inequality and count the integer solutions.

The inequality holds for

-5 \le n \le 1

. The integer solutions are

-5, -4, -3, -2, -1, 0, 1

, giving a total of

1 - (-5) + 1 = 7

integers.

The quadratic expression is non-positive between its two real roots, inclusive of the endpoints.

Solving Absolute Value Inequalities via Squaring and Quadratic Factoring

Estimated Time:1m 30s

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