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Quadratic equations and inequalities

  1. Jun 24, 2012 #1
    Well suppose for an example of an inequality,
    |x-1|-|x|+|2x+3| > 2x+4

    Well in one of its solutions we were told to apply the method of intervals, rather than taking say what; like 8 combination of signs.

    For everyone of its intervals(say -3/2 [itex]\leq[/itex]x <0) we are said that 2x+3 [itex]\geq[/itex] 0, x<0 and x-1<0.

    Is it just an intuitive outcome(guessing by the extremities of the limits) or did we do something to predict what the signs of the expressions will be?(as i was wondering if any such intervals, which violated an intuitive outcome ?)
     
  2. jcsd
  3. Jun 24, 2012 #2

    HallsofIvy

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    I'm not sure what you mean by "intuitive" but the method of "predicting" the sign is to use the definition of "absolute value": |x|= x if [itex]x\ge 0[/itex], -x if x< 0.

    Of course that says that the "change" comes when the argument of the absolute value function is 0. In order to deal with |x-1|-|x|+|2x+3| > 2x+4, I would note that x-1= 0 when x=1, x= 0 when x= 0, of course, and 2x+3= 0 when x= -3/2. Now put those in order: -3/2< 0 < 1. So if x< -3/2, it is also less than 0 and 1 and all of x-1, x and 2x+3 are negative. For x< -3/2, the inequality becomes -(x- 1)-(-x)+ (-(2x+3))> 2x+ 4. That is the same as -x+ 1+ x- 2x- 3= -2x- 2> 2x+ 4 which is, in turn, equivalent to -6> 4x or -3/2< x. Of course, if x< -3/2, "-3/2< x" can't be true so there is NO value of x, less than -3/2, satisfying that inequality.

    If [itex]-3/2\le x< 0[/itex], x and x- 1 are still negative but 2x+3 is now non-negative and so the inequality is -(x-1)-(-x)+ (2x+3)> 2x+ 4. That is the same as -x+ 1+ x+ 2x+ 3= 2x+ 4> 2x+4 which is, again, never true (they equal not "larger than"). There is no x between -3/2 and 0 satifying this inequality.

    If [itex]0\le x< 1[/itex] both 2x+3 and x are positive but x- 1 is still negative. Now the unquality is -(x-1)- x+ (2x+3)> 2x+ 3. That is the same as -x+ 1+ x+ 2x+ 3= 2x+ 4> 2x+ 3 which is always true. Every value of x, [itex]0\le x< 1[/itex] satisfies this inequality.

    Finally, if [itex]x\ge 1[/itex] all three of x-1, x, and 2x+3 are positive. Now the inequality is (x- 1)- x+ (2x+ 3)= 2x+ 2> 2x+ 3. That is never true.

    Putting all of those together, the original inequality is satified for all x, [itex]0\le x< 1[/itex] and no other x. The solution set is the interval [0, 1).
     
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