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Math Help - inequalities

  1. #1
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    inequalities

    seemingly simple problem i just cant seem to solve

    sqrt(x+5)-sqrt(8-x)<=1

    x>=5
    x<=8

    because of the square roots, i can assume that all numbers are possitive so i can open up by multiplying by the power of 2

    x+5-2*sqrt(x+5)(8-x)+8-x<=1
    therefore
    12<=2*sqrt(x+5)(8-x)
    divide by 2 and then again power of 2-to get rid of the sqrt
    36<=3x+40-x^2

    now i solve it and i get -1<=X<=4 every time but the solution is -5<=x<=4

    what am i doing wrong??
    • .
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    the answer is any x between -5 and 4, it is the right answer but i dont know how to get to it...

    i can only get from -1 to 4, but if you put any x till -5 it works, how do i solve this
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  2. #2
    o_O
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    36 \leq -x^{2} + 3x + 40
    x^{2} - 3x - 4 \leq 0
    (x - 4)(x + 1) \leq 0

    Your roots would be x = 4, x = -1 (These are the values that should theoretically satisfy the equality part of the inequality). However, if you plug in x = -1 to your original inequality, you would see that:
    \sqrt{5+(-1)} - \sqrt{8-(-1)} = \sqrt{4} - \sqrt{9} \neq 1

    So the only root you consider is x = 4. Choose a value higher than 4 and one below 4 to see which region would give you a number less than 1, keeping in mind that x is only defined over the interval -5 \leq x \leq 8
    Last edited by o_O; June 8th 2008 at 10:58 AM. Reason: something wrong in one of the latex'd expressions
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  3. #3
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    Quote Originally Posted by o_O View Post

    However, if you plug in x = -1 to your original inequality, you would see that:
    \sqrt{1+5} - \sqrt{8-1} = \sqrt{6} - \sqrt{7} \neq 1
    Sure?

    Having -5\le x\le8 and -1\le x\le4, solution set is given by [-5,8]\cap[-1,4] which gives the expected answer.
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  4. #4
    o_O
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    Ah whoops, plugged in x = 1 instead of -1.
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