How do you solve 3x^2+24x>=-413x2+24x≥−41 by graphing?

1 Answer
A. S. Adikesavan
Jan 1, 2017

x-axis with the gap (-4-sqrt(7/3), -4+sqrt(7/3))(−4−√73,−4+√73). See this gap in the graph.

Explanation:

x^2+24x=3(x+4)^2-48>=-41 to x >=-4+sqrt(7/3) and x<=-4-sqrt(7/3)x2+24x=3(x+4)2−48≥−41→x≥−4+√73andx≤−4−√73

The 2-D graph is for {(x, y)}, satisfying 3x^2+24x>=-413x2+24x≥−41.

The x-axis sans the gap

(-4-sqrt(7/3), -4+sqrt(7/3))=(-5.5275, -2.4725)(−4−√73,−4+√73)=(−5.5275,−2.4725) 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]}

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