Hello, we have : (1/x) + x < 2 (1/x) + x = 2 /*x 1 + x^2 = 2x x^2 - 2x + 1 = 0 (x-1)^2 = 0 x = 1 x < 1 Is this correct ? Becuause wolfram said result is x < 0 ?!
i think its right..
I will w8 few minutes if somebody else will find mistake. If not will close it with correct answer x < 1
whats the second line...?
linear equation ? just put "=" in place of "<"
of course i can count with "<" but "=" looks more naturally when you solve equation
k.... (1/x+x)<=|2|...was not that the condition
I dont understand you amriju ..
u cannot solve it like that, u need to consider < sign only
(1/x) + x < 2 Let x>0 , u got (x-1)^2 < 0 which is not possible because a square of a number is always positive hence x<0
only when x>0, u multiply x on both sides, and u don't flip the sign
It is not clear to me at all..
consider when x<0 so when u multiply x on both sides, u should flip the sign and u get (x-1)^2 > 0 which confirms x<0
Thanks a lot. But i still dont understand where i put mistake and how i should solve it. I dont understan english at all. I just need time to understand it
Did u know that u should flip the inequality sign when multiplying or dividing by a negative number ? ex. if x>0, -x <0 Your mistake, u cannot just replace < with = and replace it back after solving it.....
will it help if i put all steps together ?
I just know.. that if i multiply it by -1 for example.. i flip the sign of course..
but i just multiply it by x to get 1 + x^2 = 2x
Dont know if you are pointing on this step or not..
yeah, we don't know whether x is positive or negative. if positive, we need not flip the sign, if negative, we need to flip the sign. hence, we go for 2 conditions.
Ah ok.. i understand now..
but how this affect that result is x < 0 ? Iam so dumb.. sorry
(1/x) + x < 2 consider (i) x>0 (ii) x<0 u get (x-1)^2 < 0 u get (x-1)^2 < 0 NEVER true, Always true so x is not greater than 0. hence the inequality (1/x) + x < 2 is satisfied for all x<0 hence x<0 is solution
Ah, now i understand it. Thanks a lot for detailed description.. ;)
welcome ^_^
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