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Range of values

  1. May 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the range of the number k so that the equation 100^x-10^(x+1)+k=0 has two distinct positive roots

    2. Relevant equations



    3. The attempt at a solution

    I know if it says for two distinct roots only , k<25

    but now its two distinct POSITIVE roots , so how ?
     
  2. jcsd
  3. May 19, 2010 #2
    (sorry wrong, i blatantly answer without reading the question ;P)
     
    Last edited: May 19, 2010
  4. May 19, 2010 #3

    D H

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    One way to solve this problem is to find the generic solutions and then find the range of k that makes both solutions positive.

    How did you arrive at k<25? You must have done some work on this problem to arrive at that value. Show it.
     
  5. May 19, 2010 #4
    ok .

    10^(2x)-10^x . 10 + k=0

    Let 10^x be b

    b^2-10b+k=0

    Then since it says two +ve distinct roots ,

    b^2-4ac>0 , which is how i found k<25

    but the answer given is 0<k<25

    I don see where is the 0 coming , i know its sth to do with the word positive .
     
  6. May 19, 2010 #5

    D H

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    Ok so far, but your choice of b here is going to get you in trouble. Something like u would have been a much better choice:

    [tex]\aligned
    &10^{2x} - 10\,10^x + k = 0 \ \Rightarrow \\
    &u^2 - 10u + k = 0 \qquad \text{with the substitution}\ u \equiv 10^x
    \endaligned[/tex]

    This is where the choice of b will get you in trouble. This b is not the same as your variable b.

    Regarding the problem itself: Are you supposed to find the range of k that yields two distinct real solutions for x, or the range that yields two positive solutions?


    Hint: with u=10x, what values for u correspond to a real value for x? To a positive value of x?
     
  7. May 19, 2010 #6

    ok ,

    u^2 - 10u + k = 0

    then b^2-4ac>0

    (-10)^2-4k>0

    100-4k>0

    k<25 , still i get the same thing .

    And this range of values of k is supposed two distinct , positive and real solution . Is it possible ?
     
  8. May 19, 2010 #7

    D H

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    You are too hung up on the range.

    Some questions,
    • What are the solutions in terms of u?
      This is a simple quadratic equation.

    • What is x in terms of u?
      Forget about the quadratic equation in answering this question. All that matters is that u=10x.

    • How do these solutions for u translate to solutions for x?
      Use the above.

    • Does the answer to the above question always make sense?
      Just because there are two real solutions for u does not necessarily mean these translate to two real solutions for x.

    • What is the range of u that yields positive values for x?
      Once again forget about the quadratic equation in answering this question.

    • What does that translate to in terms of k[/]?
     
  9. May 19, 2010 #8


    thanks ! Or maybe i can also do it this way ,

    since i make the substitution u=10^x , u>0

    u^2-10u+k=0

    so k>0 -- 1

    Then since it has two distinct roots , b^2-4ac>0 which implies k<25 ---2

    so combining 1 and 2

    0<k<25
     
  10. May 19, 2010 #9

    D H

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    No what about two positive roots, or was a misstatement in the original post?
     
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