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Help with solving equations for x

  1. Jul 19, 2011 #1
    1. The problem statement, all variables and given/known data

    Can someone please help or guide me with these problems.
    Thanks in advance.
    less than or equal to = </=

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

    2 x^2-2x-15</= 0

    3 sin2x=sinx, 0</=x</=2pi

    4 logx+log(x-3)=1

    3. The attempt at a solution
    OK so for number 1 I do not know where to start.
    For number two I factor it and I'm in the same spot as number 1.
    As far as number 3 is concerned I do not know where to start.
    Number 4 I know that I have to condense the logarithms into one but I'm not sure what to do afterwards.
     
  2. jcsd
  3. Jul 19, 2011 #2

    vela

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    For #1, multiply the lefthand side out.

    For #2, complete the square.

    Note for #1 and #2, you could also look at the sign of each factor. If ab>0, then you must have a>0 and b>0 OR a<0 and b<0. What do you need for ab<0?

    For #3, use the trig identity sin 2x = 2 sin x cos x.

    For #4, what's the inverse of the logarithm function?
     
  4. Jul 19, 2011 #3
    #1 try ways curve method ...
     
  5. Jul 19, 2011 #4
    thank you both for the help
     
  6. Jul 19, 2011 #5

    vela

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    Did you figure them out?
     
  7. Jul 19, 2011 #6
    On number 1 can you just not set both factors equal to greater than 0 and solve for x?

    Ah completing the square! so much fun.
     
  8. Jul 19, 2011 #7

    SammyS

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    No.

    For some values of x, (x-3) and (x+3) have opposite sign. For these values, is (x-3)(x+3) positive, or is it negative?

    For other values of x, (x-3) and (x+3) have like signs. For these values, is (x-3)(x+3) positive, or is it negative?
     
  9. Jul 20, 2011 #8
    For these type of questions you can only ignore those terms to find range which are always >0 no matter what is x
     
  10. Jul 20, 2011 #9
    hmm so on 1 after multiplying it out you then complete the square to solve for x since that will give you an actual number on the right side?

    Interesting wrinkle since as long as it is a number this is how you solve these type of problems factor and then set equal to the number or 0 and solve for x.

    Though wait it has to be greater than 0 so does that not mean that both x's have to be positive? So x can be greater than 3 (from x -3) or greater than -3 from x +3

    if x is -2 then x+3 would be 1, x-3 would then be ug -5 ok so that doesn't work..interesting.
     
  11. Jul 20, 2011 #10
    #1
    completing square will give you ... x2 - 9 = 0 or x2 = 9

    now just take square root

    but don't forget [itex]\sqrt{x^2} \ = \ |x| [/itex] and not just x
     
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