How do you solve 3x^2+24x>=-41 by graphing?

Jan 1, 2017

x-axis with the gap $\left(- 4 - \sqrt{\frac{7}{3}} , - 4 + \sqrt{\frac{7}{3}}\right)$. See this gap in the graph.

Explanation:

${x}^{2} + 24 x = 3 {\left(x + 4\right)}^{2} - 48 \ge - 41 \to x \ge - 4 + \sqrt{\frac{7}{3}} \mathmr{and} x \le - 4 - \sqrt{\frac{7}{3}}$

The 2-D graph is for {(x, y)}, satisfying $3 {x}^{2} + 24 x \ge - 41$.

The x-axis sans the gap

$\left(- 4 - \sqrt{\frac{7}{3}} , - 4 + \sqrt{\frac{7}{3}}\right) = \left(- 5.5275 , - 2.4725\right)$ is 1-D solution ,

In 3-D, the gap is cylindrical, about the x-axis..

graph{3x^2+24x+41>=0 [-10, 10, -5, 5]}