Solve the inequality. $$x ^ { 4 } + x ^ { 2 } \geq 0$$
A) $( - \infty , 0 ]$
B) { }
C) $( - \infty , \infty )$
D) $( - \infty , - 1 ] \cup [ 0 , \infty )$

Question

Solve the inequality. $$x ^ { 4 } + x ^ { 2 } \geq 0$$
A) $( - \infty , 0 ]$
B) {   }
C) $( - \infty , \infty )$
D) $( - \infty , - 1 ] \cup [ 0 , \infty )$

Quizplus · Accepted Answer

The given inequality can be factored as $x^2(x^2 + 1) \geq 0$. Since $x^2$ is always non-negative for all real x, and $x^2 + 1$ is always positive (because the smallest value $x^2$ can be is 0, making $x^2 + 1 = 1$), their product is non-negative for all real x. Therefore, the solution is all real numbers, $( - \infty , \infty )$.

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

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