Question 1021512
{{{2x^2-17>0}}}
So now you want to graph the function and look for the region where the function is greater than zero (above the x-axis).
You can solve for when the function equals zero first.
{{{2x^2=17}}}
{{{x^2=(17/2)}}}
{{{x=0 +- sqrt(17/2)}}
{{{x=0 +- sqrt(34)/2)}}}
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So now choose a point in between the zeros and check whether the function is above or below the x-axis at that point.
{{{x=0}}} is convenient.
{{{2(0)^2-6>11}}}
{{{-6>11}}}
False, so in between the zeros the function is not greater than zero.
So the other regions are the where the function is greater than zero.
({{{-infinity}}},{{{-sqrt(34)/2}}})U({{{sqrt(34)/2}}},{{{infinity}}})
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*[illustration G3.JPG].