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Thread: Inequalities

  1. #1
    Newbie AnnaPorter's Avatar
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    Inequalities

    Hi! Please, help to solve the inequality :
    1) -х+3x-5>0 ;
    2) 3х-4x+8≥0 .

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  2. #2
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    Re: Solving inequalities

    WHY did you post your problem on this thread?????

    Can you solve -x^2 + 3x - 5 = 0 ?
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  3. #3
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    Re: Solving inequalities

    Quote Originally Posted by AnnaPorter View Post
    Hi! Please, help to solve the inequality :
    1) -х+3x-5>0 ;
    2) 3х-4x+8≥0 .
    @Dan It would best if this were moved to a new thread.

    @AnnaPorter. Please begin a new thread when posting a new question.

    1) Now that has no real roots bec, $(3)^2-4(-1)(-5)<0$ SEE HERE
    BUT $x=1$ we have $-(1)^2+3(1)-5<0 so there is non solution. WHY?

    2) The discriminate here is $(-4)^2-4(3)(8)<0 $ again there are no real roots SEE HERE
    but this for $x=1$ we have 3(1)^2-4(1)+8>0$ so that means it is true everywhere. WHY?
    Thanks from topsquark and AnnaPorter
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  4. #4
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    Re: Inequalities

    Another way of looking at these:
    Completing the square
    $\displaystyle -x^2+ 3x- 5= -(x^2- 3x+ (9/4- 9/4))- 5= -(x^2- 3x+ 9/4)+ 9/4- 5$
    $\displaystyle -(x- 3/2)^2+ 9/4- 20/4= -(x- 3/2)^2- 11/4$
    We can think of this as -11/4 minus a square so it is never larger than -11/4 so never larger than 0.

    $\displaystyle 3x^2- 4x+ 8= 3(x^2- (4/3)x+ (4/9- 4/9))+ 8= 3(x^2- (4/3)x+ 4/9)- 4/3+ 8$
    $\displaystyle 3(x+ 2/3)^2- 4/3+ 24/3= 3(x+2/3)^2+ 5$

    That is 5 plus a square to is never less than 5 and so never less than 0.
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