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.

ant101 Dec 27, 2018

#2**+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

CPhill Dec 27, 2018