Solve: $x ^ { 2 } - 4 x - 12 \leq 0$
A) $[ - 2,6 ]$
B) $[ - 6,2 ]$
C) $( - \infty , - 2 ] \cup [ 6 , \infty )$
D) $( - \infty , - 6 ] \cup [ 2 , \infty )$

Question

Quizplus · Accepted Answer

To solve the inequality $x^2 - 4x - 12 \leq 0$, first factor the quadratic equation if possible. The given equation can be factored as $(x-6)(x+2) \leq 0$. The roots of the equation are $x = 6$ and $x = -2$. To determine the intervals where the inequality holds, consider the sign of the product in each interval determined by the roots: $x < -2$, $-2 < x < 6$, and $x > 6$. Testing values from each interval in the inequality shows that the inequality holds for $-2 \leq x \leq 6$, which corresponds to the interval $[ - 2,6 ]$.

Solve x2−4x−12≤0x ^ { 2 } - 4 x - 12 \leq 0x2−4x−12≤0

Solve $x ^ { 2 } - 4 x - 12 \leq 0$