Question 1200712
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(1) Rewrite the inequality in standard form with "0" on the right -- just as we would do if we were solving a quadratic equation:<br>
{{{3(x^2-1)>-8x}}} --> {{{3x^2+8x-3>0}}}<br>
(2) Factor the quadratic expression:<br>
{{{3x^2+8x-3>0}}} --> {{{(3x-1)(x+3)>0}}}<br>
(3) Determine the solution set.<br>
The expression is EQUAL to 0 at x=-3 and x=1/3.  The sign of the evaluated expression can only change at those two values of x.<br>
So, because the inequality is a strict inequality (it can't be equal to zero), we have three intervals of interest: (-infinity,-3), (-3,1/3), and (1/3,infinity).  There are two basic ways to determine which of those intervals are part of the solution set.<br>
One elementary way is to choose a test value in each of the intervals and find the intervals on which the inequality is satisfied.  I will leave that to you.<br>
Another elementary way is to know that the graph of the quadratic expression is an upward-opening parabola, so the expression is negative only between x=-3 and x=1/3.  We then know on which intervals the expression is greater than zero.<br>
ANSWER: (-infinity,-3) U (1/3,infinity)<br>