Solve the inequality. Write the solution set in interval notation.
-$$\frac { ( x - 2 ) ^ { 2 } } { x ^ { 2 } - 16 } 0$$
A) $( - \infty , - 4 ) \cup ( 4 , \infty )$
B) $( - 4,2 ) \cup ( 4 , \infty )$
C) $( - \infty , - 4 ) \cup ( 2,4 )$
D) $( - 4,2 ) \cup ( 2,4 )$

Question

Solve the inequality. Write the solution set in interval notation.
-$$\frac { ( x - 2 ) ^ { 2 } } { x ^ { 2 } - 16 } > 0$$
A) $( - \infty , - 4 ) \cup ( 4 , \infty )$
B) $( - 4,2 ) \cup ( 4 , \infty )$
C) $( - \infty , - 4 ) \cup ( 2,4 )$
D) $( - 4,2 ) \cup ( 2,4 )$

Quizplus · Accepted Answer

The inequality $\frac { ( x - 2 ) ^ { 2 } } { x ^ { 2 } - 16 } > 0$ is positive when the numerator and denominator have the same sign and not equal to zero. The denominator $x^2 - 16 = (x-4)(x+4)$ is zero at $x = 4$ and $x = -4$, creating critical points that divide the number line into intervals. Testing intervals around these points, we find the function is positive outside the interval $(-4, 4)$, hence the solution is $( - \infty , - 4 ) \cup ( 4 , \infty )$.

Solve the Inequality (x−2)2x2−16>0\frac { ( x - 2 ) ^ { 2 } } { x ^ { 2 } - 16 } > 0x2−16(x−2)2​>0

Solve the Inequality $\frac { ( x - 2 ) ^ { 2 } } { x ^ { 2 } - 16 } > 0$