Question 98471
{{{3(t-1)(t-3) > 0}}} Start with the given inequality


{{{3(t-1)(t-3) = 0}}} Set the left side equal to zero



Now set each factor equal to zero:


{{{t-3=0}}} or {{{t-1=0}}}


{{{t=3}}} or {{{t=1}}}  Now solve for t in each case



So our solutions are {{{t=3}}} or {{{t=1}}}



These values make the entire inequality equal to zero. So that means that at {{{t=3}}} or {{{t=1}}} the graph will go from positive to negative (or vice versa). So we must use these points to test the inequality



So now pick any value less than {{{t=1}}} and test the inequality {{{3(t-1)(t-3) > 0}}}. So lets pick t=0



{{{3(0-1)(0-3) > 0}}} Plug in t=0


{{{3(-1)(-3) > 0}}} Subtract


{{{9> 0}}} Multiply


Since t=0 makes the inequality true, that means everything less than 1 will make the inequality true. So one part of our answer is {{{t<1}}}.



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Now pick any point in between  {{{t=1}}}  and {{{t=3}}}. Let's pick t=2





{{{3(2-1)(2-3) > 0}}} Plug in t=2


{{{3(1)(-1) > 0}}} Subtract


{{{-3> 0}}} Multiply


Since t=2 makes the inequality false, that means the possible answer {{{t>1}}} false. So this shows why {{{t>1}}} is not an answer.


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Now pick any point greater than {{{t=3}}}. Let's pick t=4




{{{3(4-1)(4-3) > 0}}} Plug in t=0


{{{3(3)(1) > 0}}} Subtract


{{{9> 0}}} Multiply


Since t=4 makes the inequality true, that means one part of our answer is {{{t>3}}}



That last part is somewhat unnecessary since we already eliminated everything else. However, it is important to see everything that is going on





So our solution is {{{t<1}}} or {{{t>3}}}




Notice if we graph the equation we get




{{{ graph( 500, 500, -10, 10, -10, 10, 3(x-1)(x-3)) }}} Graph of {{{3(t-1)(t-3)}}}



and we can clearly see that everything to the left of 1 is greater than zero. Also, we can see that everything to the right of 3 is greater than zero. So our answer is verified.