How many integer values of x satisfy the inequality |x^2 - 5x| \le 6? — GMAT

How many integer values of $x$ satisfy the inequality $|x^2 - 5x| \le 6$ ?

Answer: 8

The correct answer is 8.

To solve

|x^2 - 5x| \le 6

, express it as

-6 \le x^2 - 5x \le 6

. Solving

x^2 - 5x - 6 \le 0

gives

[-1, 6]

, while solving

x^2 - 5x + 6 \ge 0

gives

(-\infty, 2] \cup [3, \infty)

. Taking their intersection yields the solution set

[-1, 2] \cup [3, 6]

. The integers contained in this set are

-1, 0, 1, 2, 3, 4, 5, 6

, which gives 8 distinct integer values.

Rewrite the absolute value inequality

-6 \le x^2 - 5x \le 6

For any real expression

A

and constant

k \ge 0

|A| \le k

is equivalent to

-k \le A \le k

Solve the upper bound condition

x^2 - 5x \le 6

-1 \le x \le 6

Subtract 6 from both sides to obtain

x^2 - 5x - 6 \le 0

. Factoring gives

(x - 6)(x + 1) \le 0

Solve the lower bound condition

x^2 - 5x \ge -6

x \le 2

x \ge 3

Add 6 to both sides to obtain

x^2 - 5x + 6 \ge 0

. Factoring gives

(x - 2)(x - 3) \ge 0

Combine the solution sets

[-1, 2] \cup [3, 6]

The intersection of

[-1, 6]

with

(-\infty, 2] \cup [3, \infty)

is the set of intervals

[-1, 2]

and

[3, 6]

Count all integer solutions in the combined set

8 integer values

The integer values in

[-1, 2]

are

-1, 0, 1, 2

(4 integers), and in

[3, 6]

are

3, 4, 5, 6

(4 integers), totaling

4 + 4 = 8

integers.

Solving quadratic absolute value inequalities using compound inequality decomposition.

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