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avatar+874 

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
 #1
avatar+5226 
+1

editor keeps mangling my answer.

 

The final answer is

 

\(\dfrac{t(2t-3)}{4t-2}\leq 0 \Rightarrow t \in (\infty,0] \bigcup \left(\dfrac 1 2, \dfrac 2 3\right]\)

.
 Dec 27, 2018
edited by Rom  Dec 27, 2018
edited by Rom  Dec 27, 2018
 #2
avatar+101870 
+1

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

 

 

cool cool cool

 Dec 27, 2018
 #3
avatar+874 
0

Yes! Thanks, everyone!

 Dec 28, 2018

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