How many integer values of x satisfy both |x^2 - 10| \le 6 and |x - 1| > 2? — GMAT

How many integer values of $x$ satisfy both $|x^2 - 10| \le 6$ and $|x - 1| > 2$ ?

Answer: 4

4 integer values satisfy both inequalities:

-4

-3

-2

, and

4

Solving

|x^2 - 10| \le 6

gives

4 \le x^2 \le 16

, which yields 6 integer solutions:

\{-4, -3, -2, 2, 3, 4\}

. Applying the condition

|x - 1| > 2

requires

x < -1

x > 3

. Only

-4, -3, -2

, and

4

satisfy both conditions, giving exactly 4 valid integer values.

Solve the quadratic absolute value inequality

|x^2 - 10| \le 6

-6 \le x^2 - 10 \le 6 \implies 4 \le x^2 \le 16

Removing the absolute value bars gives the bounded inequality

-6 \le x^2 - 10 \le 6

. Adding 10 across all parts isolates

x^2

Identify all candidate integer values of

x

for which

4 \le x^2 \le 16

x \in \{-4, -3, -2, 2, 3, 4\}

Taking the square root yields

-4 \le x \le -2

2 \le x \le 4

. The integer solutions are

-4, -3, -2, 2, 3, 4

Solve the linear absolute value inequality

|x - 1| > 2

x > 3

x < -1

An absolute value distance greater than 2 expands into two disjunctive linear inequalities:

x - 1 > 2

x - 1 < -2

Intersect the candidate integer set

\{-4, -3, -2, 2, 3, 4\}

with the domain

x < -1

x > 3

The valid integers are

-4, -3, -2,

and

4

. The values

2

and

3

fail the inequality since they fall inside the excluded interval

[-1, 3]

Testing each candidate shows that

-4, -3, -2

are strictly less than

-1

, and

4

is strictly greater than

3

Solving compound absolute value inequalities and finding integer solution set intersections.

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