Hey! I Wwas wondering if any of you coul dhelp me with a math problem im having im a bit stuck as to what to do with this modulusus inequaliy! |x-3| < 4 does anybody know what do do with that to find x ? thanks!
Been a while since I've done something like that, but as I remember, you solve for |x-3| = 4 to get 2 values. You should get 7 and -1. These are your outer bounds. Then plot those points on a number line: <------(-1)--(0)----------------(7)--------> Then you need to test points on either side of the bounds, so substitute a value less than -1 in the original equation, substitute a value in between -1 and 7, and substitute a value greater than 7. so |-2-3| < 4 That comes up "True" |5-3| < 4 That is also true. |8-3| < 4 This is false! So your answer should be (if I did this right ) x<7. Anyone want to see If I did this right? Hope this helps. ~David
oh i see thanks alot and mrhaz, that was well unneccessary this is not a span trap you know, if you want a spam trap go here http://forums.bit-tech.net/showthread.php?s=&threadid=29785
As long as Sqrt() is defined as the positive square root As Madyeti pointed out you can do this one by inspection, though the fool proof method is to change the < to an =, square both sides (as DaSuperFly said), work out the possible values of x and then probe the original equation with the values to see how it behaves so square both side, (x-3)(x-3) = 16 x^2 - 6x - 7 = 0 (x+1)(x-7) = 0 so x is -1 or 7, then go as Madyeti said 'cept he made a mistake when he put x=-2 into the equation |-2-3| < 4 gives 5<4 which is false... so -1<x<7 is the answer I think Rob.
