dave13 wrote:
VeritasKarishma many thanks for explanation
one question, if i multiply whole inequity \((-x)(2x-5)(x-6)>0\) by -1
then i get \((x) (-2x+5) (-x+6)<0\)
how is it possible that signs inside of brackets didn`t change in your case after multiplying by -1
Think about this:
3 * 5 > 0
Multiply both sides by 2 to get:
2 * 3 * 5 > 0
Does it become 2*3 * 2*5 > 0? No.
They are all factors. When you multiply by 2, one more factor comes in.
You can write it as 2*3 * 5 > 0 or as 3 * 2*5 > 0 but not as 2*3 * 2*5 > 0
Similarly,
(-x) (2x-5)(x-6)>0 has three factors. When you multiply by -1, only one term needs to get multiplied by -1, not all.
You are confusing it with
(3 + 5) > 0
Now if you multiply by 2, you will multiply both terms with 2. This happens only when you have a + or a - sign.
in your last step you have \(2(x – 0)(x – 5/2)(x – 6) < 0\) you have \(2(x-0)\) may i know how you got it, i thought maybe you factored out 2
but if it is so why are you subtracting 0 and not adding it though yes adding or subtracting zero doesnt change solution ....but still kinda curious