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Solution related to certain condition in quadratic equation

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data
    For a constant k, consider the function f(x) = x2 + kx + k2 - 2k - 4
    Find the range of k for which f(x) = 0 has one solution between 0 and 1 and the other solution between 1 and 2


    2. Relevant equations
    quadratic formula
    discriminant


    3. The attempt at a solution
    x2 + kx + k2 - 2k - 4 = 0

    Using quadratic formula:
    [tex]x=\frac{-k±\sqrt{-3k^2+8k+16}}{2}[/tex]

    Then I set:
    [tex]0 < \frac{-k-\sqrt{-3k^2+8k+16}}{2} < 1[/tex] and [tex]1 < \frac{-k+\sqrt{-3k^2+8k+16}}{2} < 2[/tex]

    Am I on the right track? I think solving that inequality will take a lot of work... Is there other method?

    Thanks
     
  2. jcsd
  3. May 12, 2012 #2
    Your method is correct, though very tedious. Another thing you can try is see how the graph of the quadratic is.

    Meaning, the curve f(x) intersects x axis(has 2 roots), what condition does this give you for the discriminant?
    Then, for a root to be in (0,1) and the other to be in (1,2) what should the nature of product of f(0) and f(2) be? Will this be a necessary -and- sufficient condition?

    Drawing a rough graph will help :smile:
     
  4. May 12, 2012 #3

    ehild

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    Try to use the relation between coefficients and roots of the quadratic equation.

    ehild
     
    Last edited: May 13, 2012
  5. May 13, 2012 #4
    This is actually an easier method to understand, than I suggested :smile:
     
  6. May 13, 2012 #5
    I am not sure what you mean by "relation between coefficients and roots of the quadratic equation". Are you talking about Vieta's formula?

    If yes, I don't see how Vieta's formula can form inequality to find range of k...

    If no, then I don't what your hint means...:redface:
     
  7. May 13, 2012 #6
    Yes, the Vieta formula. Consider the roots to be A and B.

    A + B = ??

    A*B = ??

    Hint : you already know the lowest possible value for your roots is in (0,1) and the highest is in (1,2) How can you get an interval of sum of roots and product from this?
     
  8. May 13, 2012 #7

    ehild

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    It was Vieta! I could not remember the name.:blushing:
    One condition comes also from the discriminant -3k2+8k+16. It can not be negative.
    So you have three condition for k...

    ehild
     
  9. May 13, 2012 #8
    OK let me try :smile:

    Let the roots of the equation f(x) = 0 be A and B. So:

    A + B = -k and 1 < A + B < 3, then 1 < -k < 3 leads to -3 < k < -1

    A.B = k2 - 2k - 4 and 0 < AB < 2, then 0 < k2 - 2k - 4 < 2
    1. 0 < k2 - 2k - 4 --> solving this, I got : k < 1 - √5 or k > 1 + √5

    2. k2 - 2k - 4 < 2 --> solving this, I got : 1 - √7 < k < 1 + √7

    So the range of k for 0 < k2 - 2k - 4 < 2 is 1 - √7 < k < 1 - √5 or 1 + √5 < k < 1 + √7


    The intersection of -3 < k < -1 and (1 - √7 < k < 1 - √5 or 1 + √5 < k < 1 + √7) is 1 - √7 < k < 1 - √5

    But if I remember correctly, the solution is (1 - √13) / 2 < k < 1 - √5

    Where is my mistake?

    Thanks
     
  10. May 13, 2012 #9
    There you go.
     
  11. May 13, 2012 #10
    -3k2+8k+16 ≥ 0

    Solving that, I got : -4/3 ≤ k ≤ 4 and that leads to the same solution as previous: 1 - √7 < k < 1 - √5 :confused:
     
  12. May 13, 2012 #11

    ehild

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    1-√7=-1.645. -4/3 = -1.333. 1-√5=-1.236. So what is the domain for k?

    ehild
     
  13. May 13, 2012 #12
    Edit again.

    I just found a fatal problem with this method of using only Vieta formula and discriminant. In your cases you also consider the cases when both A and B are between (0.5,1) since addition and multiplication of these two numbers will satisfy the inequality you get. To rectify this, think about what the nature of f(1) should be? Refer my post #2 for that.
     
    Last edited: May 13, 2012
  14. May 13, 2012 #13
    Oops sorry my bad. I thought 1 - √7 > -4/3

    So, the domain for k is -4/3 < k < 1 - √5

    But still it's not the same as the answer key, which is (1 - √13) / 2 < k < 1 - √5

    How can we get term (1 - √13) / 2 from above working?
     
  15. May 13, 2012 #14
    Please see post #12 again.
     
  16. May 13, 2012 #15
    Sorry, when I posted, your post #12 consists only ehild replied :smile:

    I found the answer thanks to your hint. But how can you know we need to eliminate the case [0.5 , 1]? And how can we be sure that there are no others fatal problems such as [0.5 , 1]?

    If there was no answer key, I will answer -4/3 < k < 1 - √5 confidently :tongue:
     
  17. May 13, 2012 #16
    To see whether there is a problem, the best way is to see if you can find any example that will contradict your answer :wink:

    The method I explained in post #2 is foolproof. You only need to be cautious. :smile:
     
  18. May 13, 2012 #17

    ehild

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    I do not understand. Just the Vieta formulae were not enough. But we added the condition of the non-negativity of the discriminant with k.
    It is not required that any pair from (0,1) and (1,2) be solutions.

    The official solution is (1 - √13) / 2 < k < 1 - √5. Approximately it means -1.30<k<-1.23

    We got -1.33<k<-1.23. And k=-1.32 is still appropriate.

    ehild
     
  19. May 13, 2012 #18
    Sometimes it is pretty hard to find any example that will contradict the answer, such as this one, because -4/3 is not so much differs from (1 - √13) / 2 :biggrin:

    But I'll try to keep that in mind :smile:

    By the way, I think using f(0) > 0 and f(1) < 0 will give correct result and it is faster ?? I got this idea from your post :smile:
     
  20. May 13, 2012 #19
    I am not sure I get what you mean but substituting k = -1.32 to f(x) will result that the solution does not lie on interval (1,2)
     
  21. May 13, 2012 #20

    ehild

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    Sorry, I miscalculated something. Infinitum is right.
    But it was a nice problem....

    ehild
     
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