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# Real Values

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Find all real values of $$t$$ which satisfy $$\frac{t(2t-3)}{4t-2} \le 0.$$

I got $$\:t\le \:0\quad \mathrm{or}\quad \frac{1}{2} , but how would I express this. Dec 27, 2018 ### 3+0 Answers #1 +6250 +1 editor keeps mangling my answer. The final answer is \(\dfrac{t(2t-3)}{4t-2}\leq 0 \Rightarrow t \in (\infty,0] \bigcup \left(\dfrac 1 2, \dfrac 2 3\right]$$

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Dec 27, 2018
edited by Rom  Dec 27, 2018
edited by Rom  Dec 27, 2018
#2
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Here's one way to do this

Just  set this = 0   and   multiply through by  4t - 2   and we get that

t (3t - 2)  =  0

Setting each factor to  0    and solving, we get that

t = 0      or     t = 2/3

And since the denominator of the original fraction cannot = 1/2....we have these  possible intervals

(-inf, 0]  ( 0, 1/2)  (1/2, 2/3 ]  and ( 2/3, inf )

Picking a point in each interval  shows that the intervals   (-inf, 0 ]  and (1/2, 2/3]  solve the inequality

Dec 27, 2018
#3
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Yes! Thanks, everyone!

Dec 28, 2018