Solve the inequality. Write the solution set in interval notation.
-$$x ^ { 4 } - 13 x ^ { 2 } + 36 \geq 0$$
A) $( - 3 , - 2 ) \cup ( 2,3 )$
B) $( - \infty , - 3 ] \cup [ - 2,2 ] \cup [ 3 , \infty )$
C) $( - \infty , - 3 ) \cup ( - 2,2 ) \cup ( 3 , \infty )$
D) $[ - 3 , - 2 ] \cup [ 2,3 ]$

Question

Quizplus · Accepted Answer

To solve the inequality $x^4 - 13x^2 + 36 \geq 0$, we first factor the left-hand side. Let $y = x^2$, then the inequality becomes $y^2 - 13y + 36 \geq 0$. Factoring this gives $(y - 9)(y - 4) \geq 0$, which means $y \geq 9$ or $y \leq 4$. Substituting back $x^2$ for $y$, we get $x^2 \geq 9$ or $x^2 \leq 4$. This translates to $x \leq -3$, $x \geq 3$, or $-2 \leq x \leq 2$, which in interval notation is $(-\infty, -3] \cup [-2, 2] \cup [3, \infty)$.

Solve the Inequality x4−13x2+36≥0x ^ { 4 } - 13 x ^ { 2 } + 36 \geq 0x4−13x2+36≥0

Solve the Inequality $x ^ { 4 } - 13 x ^ { 2 } + 36 \geq 0$