Solve each inequality. Express in interval notation. 2x^2+6x less than/equal to 0.

between its roots id hope

2x(x+3); so between -3 and 0

\[2x^2+6x \le 0\]\[2x(x+3)\le0\]\[x \le 0\]\[x \le -3\]since the limiting factor is the second one:\[x≤−3\rightarrow(-\infty,-3]\]

So you don't do anything with the \[x \le 0\]

no because we know that x cannot be greater than -3. All the above says is that x is AT MOST 0. The other statement says that x is AT MOST -3. So the fact that\[x \le -3\]implies that \[x \le 0\]

ok. Thanks

spose x = -10 2(-10)^2+6(-10) 2(100) - 60 is not less than zero so x <-3 doesnt work

the interval should be: [-3,0]

lets try x=-4 2(-4)^2+6(-4) 2(16)-24 32-24 is greater than 0

graph wise, a parabola curves so that at best there is only a distinct section that is below the x axis

to say that it is below the x axis from -3 to -infinity is not accurate to be polite :)

http://www.wolframalpha.com/input/?i=2x^2%2B6x as you can see from the graph, it is equal to or less than zero from -3 to 0

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