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Simple Quadratic

  1. Apr 1, 2007 #1
    1. The problem statement, all variables and given/known data

    Solve for all real x: (x^2-6x+9)^(x^2-4)=1

    2. Relevant equations

    3. The attempt at a solution

    I raised both sides by 1/x^2-4, so got x^2-6x+9=1, solved it to get x=2,4.
    It's a question from a college enterance exam sample paper, so I wondered if it could be that simple. Is my Solution correct?
    Last edited: Apr 1, 2007
  2. jcsd
  3. Apr 1, 2007 #2


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    I'm assuming you didn't mean to put '^' between the two factors. In which case just remember a*b=0 if and only if a=0 or b=0.
  4. Apr 1, 2007 #3
    It's x^2-6x+9 riased to the power x^2-4
  5. Apr 1, 2007 #4
    What's [tex](x^2-6x+9)^0[/tex] ?
  6. Apr 1, 2007 #5
    1. How will that help?
  7. Apr 1, 2007 #6
    anything to power 0 is 1
  8. Apr 1, 2007 #7
    so? Ho is it relevant to the ques?
  9. Apr 1, 2007 #8
    Well anything raised to the 0 power is going to equal 1. So, how would you get [tex](x^2-6x+9)^{x^2- 4}[/tex] to be [tex](x^2-6x+9)^0[/tex]?
  10. Apr 1, 2007 #9
    OK I get your point. x=2,-2. But what is wrong with my solution?
  11. Apr 1, 2007 #10
    Well then if you can turn the exponent to 0, then wouldn't the equation also be satisfied when x=-2? (You already have 2 as a solution) Do you see what I'm getting at?

    If you don't, graph the equation (x^2-6x+9)^(x^2-4)-1=0 and look at the x-intercepts.
  12. Apr 1, 2007 #11
    Ok if I take log on both sides then I get (x^2-4)log(x^2-6x+9)=log1=0.
    then i get x^2-4=0, for which i get x=2,-2 or x^2-6x+9=1 for which i get x=2,4. so is the answer x=2,-2 or x=2,4?
  13. Apr 1, 2007 #12
    Not quite, the answer is going to be x=-2,2,4.
  14. Apr 1, 2007 #13
    Why not or?
    I couldn't graph the func. still haven't learnt enough.
  15. Apr 1, 2007 #14
    It's going to be x=-2,2,4 because if x=-2 or 2 then the exponent becomes 0, and anything raised to the 0 power will equal 1. x=4 is a solution because:

    [tex](x^2-6x+9)^{x^2- 4}=1[/tex]
    [tex]((x^2-6x+9)^{x^2- 4})^{1/x^2- 4}=1^{1/x^2- 4}[/tex]
    [tex] x=2,4[/tex]

    So, 4 is also a solution
  16. Apr 1, 2007 #15
    Its simple. You get 1 on RHS iff the exponent is 0 OR/AND the base is itself 1.
    So, [itex]\ x^2-6x-9=1 \ \mbox{OR/AND} \ x^2-4=0 [/itex]
  17. Apr 2, 2007 #16
    Thanks for the help.
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