Question

The set of all real values of $x$ for which $(x^2 - |x+9| + x)>0$ is:

• A. $(-\infty,-9)\cup(3,\infty)$
• B. $(-\infty,-3)\cup(9,\infty)$
• C. $(-\infty,-3)\cup(3,\infty)$
• D. $(-9,-3)\cup(3,9)$

Hint:

When solving inequalities involving absolute values, always split into cases based on the sign of the expression inside the absolute value. Analyze each case separately, then combine intervals carefully by intersection and union.

Answer 1

The Correct Option is C

Approach: Skip the casework and use the options as a sieve. Pick one boundary value and a couple of test points; whichever option's set agrees with the true sign of the expression at those points must be the answer.

Let $E(x) = x^2 - |x+9| + x$. We need $E(x) > 0$.

Test $x = 0$ (a point all options near the middle care about): $E(0) = 0 - 9 + 0 = -9 < 0$. So $0$ must be excluded. Options 1, 3, 4 all exclude $0$; option 2 excludes the interval $[-3,9]$ which contains $0$ — keep watching.

Test $x = -5$: $E(-5) = 25 - |4| - 5 = 25 - 4 - 5 = 16 > 0$. So $-5$ must be included. Option 1 excludes $[-9,3]$ (rejects $-5$) — out. Option 2 includes $-5$ (its excluded zone is $[-3,9]$) — survives. Option 3 includes $-5$ — survives. Option 4 excludes $-5$ — out.

Decide between options 2 and 3 using $x = 5$: $E(5) = 25 - 14 + 5 = 16 > 0$, so $5$ must be included. Option 2 excludes $5$ (its gap is $(-3,9)$) — out. Option 3 includes $5$ — it survives every test.

Final answer: $(-\infty,-3) \cup (3,\infty)$ — option 3.