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Solving inequality

  1. Jul 20, 2008 #1
    1. The problem statement, all variables and given/known data

    sqrt(x-1)+sqrt(x-3)+8>0

    2. Relevant equations



    3. The attempt at a solution
    I thought I would square everything to get rid of the root signs...leaving me with
    (x-1)+(x-3)+64>0
    2x+60>0
    x>30
    ss:{xE[30,Infinity)}

    However, the answer is supposed to be..
    [3,infinity)
    Please tell me what I am missing.
     
  2. jcsd
  3. Jul 20, 2008 #2
    This is your problem right?

    [tex]\sqrt{x-1}+\sqrt{x-3}+8 > 0[/tex]

    If you "square both sides" you will not get: (x-1)+(x-3)+64>0 !

    Look at this web page. The same rules shown there will apply if you square

    [tex]\sqrt{x-1}+\sqrt{x-3}+8[/tex]

    Only, think of [tex]\sqrt{x-1}=a[/tex] [tex]\sqrt{x-3}=b[/tex] and 8=c.

    As you can see, when you square a trinomial you may end up with a big mess.

    You can solve this problem by thinking about it for a bit. For this to be true:

    [tex]\sqrt{x-1}+\sqrt{x-3}+8 > 0[/tex]

    we know that

    [tex]\sqrt{x-1}+\sqrt{x-3}> -8[/tex]

    Well, when are square roots ever negative?
    So, what restrictions should you place on X?
     
  4. Jul 20, 2008 #3
    Okay, I didn't recognize it as a trinomial. If x is less than 3 the equation won't work, right?
    All I have to do is recognize what the minumum x must be so that I am not taking a sqrt of a negative number. Is that correct?

    If the equation was sqrt(x-5)+sqrt(x-9)+14>0 would I be correct in stating that xE[9,infinity)
    because then I am not taking the sqrt of a negative number. Is that all I have to do for this kind of problem?
     
  5. Jul 26, 2008 #4

    dynamicsolo

    User Avatar
    Homework Helper

    I'm surprised you hadn't gotten a reply by now...

    Yes, you are correct. In this particular situation, since the square root operation only gives results that are positive or zero, the left-hand side of

    [tex]
    \sqrt{x-1}+\sqrt{x-3}+8 > 0
    [/tex]

    couldn't possibly be less than 8. So plainly the left-hand side is always greater than zero.

    So what is the only problem you could run into? That the terms on the left-hand side be undefined. The first term is only defined for [tex]x - 1 \geq 0[/tex] and the second term, for [tex]x - 3 \geq 0[/tex]. The intersection of those sets is [tex]x \geq 3[/tex]. You can make a similar argument for the inequality you suggested, for which you found [tex]x \geq 9[/tex].
     
  6. Jul 27, 2008 #5
    sqrt(x-1) + sqrt(x-3) + 8 > 0

    Well it will always be greater than 0, in fact always greater than or equal to 8 atleast, since the roots cannot be negative

    now just consider the domain x >=3 ... because of the roots again
    And that's your answer x E [3, infinity)
     
  7. Jul 27, 2008 #6
    Just find the domain and you've got your answer. You don't need to square or move anything around.
     
  8. Jul 27, 2008 #7
    Thank you.
     
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