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Real Roots of Exponential Equation (Involves Quadratic)

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Hi. I actually understand most of this question, but not the parts in red.

    Question.

    [PLAIN]http://img703.imageshack.us/img703/7237/2008testhphysf.jpg [Broken]

    If above doesn't load, please go to http://img703.imageshack.us/img703/7237/2008testhphysf.jpg [Broken]

    2. Relevant equations

    The roots of a quadratic equation are real when [tex] b^2 - 4ac \geq 0 [/tex]


    3. The attempt at a solution

    Because we want real solutions, we have...

    [tex] b^2 - 4ac \geq 0 [/tex] so [tex] 9 + 4k \geq 0 => \sqrt{9 + 4k} \geq 0 [/tex] .

    But [tex] y^2 - 3y + k = 0 [/tex] has two solutions...

    [tex] y_1 = \frac{1}{2}(3 - \sqrt{9 + 4k} [/tex]

    [tex] y_2 = \frac{1}{2}(3 + \sqrt{9 + 4k} [/tex]

    I get that [tex] 9 + 4k \geq 0 [/tex] means [tex] y_2 is true [/tex], but what about [tex] y_1 [/tex]?

    The solution cares about "the larger root". Ie y_2. But why doesn't it care about y_1? I mean, both y_1, y_2 are solutions?

    Thank you.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 18, 2011 #2

    dynamicsolo

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    Your attachment isn't coming through.
     
  4. Sep 18, 2011 #3

    gb7nash

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    Context? Sometimes, when you're dealing with time problems, you'll get a positive and negative root, but since a negative time isn't valid, you'll just take the positive root.

    However, if you're just given the equation:

    [tex]y^2 -3y + k = 0[/tex]

    You are correct in saying that y1 and y2 are distinct real roots (assuming that 9+4k > 0).
     
  5. Sep 18, 2011 #4
    Hi. I think I fixed my original post. The picture loads now.

    gb7nash. Thanks for your answer. But this question and the test don't give any context for [tex] x [/tex] so I still don't see why the solution doesn't care about the negative root?
     
  6. Sep 18, 2011 #5

    gb7nash

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    Ok, I can see the image now.

    Originally, we substitute y = 3x and we want to solve for y. Think about it for a second. Is there any way y can be negative? Is 3 raised to any power of x a negative number?

    Knowing this now...
     
  7. Sep 18, 2011 #6
    Hi gb7nash. Thanks for your answer.

    Actually, I understand why [tex] 3^x > 0 [/tex] for all x.

    But my question is on how [tex] y_1 = \frac{1}{2}(3 - \sqrt{9 + 4k}) [/tex] goes with the above.

    We don't know [tex] 3 - \sqrt{9 + 4k} > 0 [/tex]

    We only know [tex] \sqrt{9 + 4k} \geq 0 [/tex]?
     
  8. Sep 18, 2011 #7

    gb7nash

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    You're exactly right. Depending on what value you choose for k, y1 may be positive or negative. For instance, for k = 10 you'll obtain a negative value for y1 (which is nonsense, since y cannot equal a negative number). For k = 0, you'll obtain a positive value for y1.

    Since this is the case, we can't always rely on y1 to give a valid root, since there is the possibility of it being negative. However, y2 will always give a positive number (for k >= -9/4) and a valid root.
     
  9. Sep 18, 2011 #8
    Hi gb7nash. Thanks for your answer.

    So are you saying the only final answer to this question is just y2, GIVEN [tex] \sqrt{9 + 4k} \geq 0 [/tex]?

    I should toss out y1, since [tex] 9 + 4k \geq 0 [/tex] for this problem, because like we just talked about, y1 isn't always > 0 if [tex] 9 + 4k \geq 0 [/tex]?

    So just to make sure, y2 = 3x, GIVEN [tex] \sqrt{9 + 4k} \geq 0 [/tex]. AND it is WRONG to write y = 3x as final answer, if [tex] 9 + 4k \geq 0 [/tex].
     
    Last edited: Sep 18, 2011
  10. Sep 18, 2011 #9

    gb7nash

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    No. A certain range of k makes y1 valid. What must k be to make y1 positive?
     
  11. Sep 18, 2011 #10
    Hi gb7nash. Thanks for your answer.

    I edited my second last post because I actually didn't specify requirement [tex] 9 + 4k \geq 0 [/tex] for this problem. Now did I get everything right?

    But to answer your question, y1 > 0 if [tex] 3 - \sqrt{9 + 4k} > 0 <=> \sqrt{9 + 4k} < 3 <=> k < 0 [/tex]. But this isn't what we want for this question. Because if k < -9/4, original equation won't have real roots.
     
    Last edited: Sep 18, 2011
  12. Sep 18, 2011 #11

    gb7nash

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    No, y1 > 0 if:

    [tex]\sqrt{9+4k} < 3[/tex]

    Solving for k, what do you get?

    Also, you want 9+4k > 0 in order to obtain a real root. Solve for k again.

    Combining these, what must k be between to have two roots?
     
  13. Sep 18, 2011 #12
    Hi gb7nash. Thanks for your answer.

    I actually meant to write [tex] 3 - \sqrt{9 + 4k} > 0 [/tex] so I've fixed this.

    Because 3x = y > 0, the asked exponential question has AT LEAST two roots if

    k < 0 and k > -9/4 <=> -9/4 < k < 0.

    But I just want to make sure. The question asks "one or more real solutions", so isn't [tex] k \geq -9/4 [/tex] enough?

    This'd mean that it's enough to have y2 as the solution? y1 isn't necessary? Because "one" real solution's enough? Or is the answer wrong?
     
  14. Sep 18, 2011 #13

    gb7nash

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    Yes. If k >= -9/4, we can guarantee that we'll have one or two real roots (depending on if k is between -9/4 and 0, or greater than 0). I interpret one or more to mean that they want both cases covered, so k >= -9/4 gives you what you need.
     
  15. Sep 18, 2011 #14
    Hi gb7nash. Thank you so much for your answers.

    So again, just to make sure, for only one real root, then

    only y2 [tex] = \frac{1}{2}(3 + \sqrt{9 + 4k} [/tex] is a solution

    so only y2 = 3x.

    y1 [tex] = \frac{1}{2}(3 - \sqrt{9 + 4k} [/tex] is NOT a solution. I just want to make sure because I was confused about this.
     
  16. Sep 19, 2011 #15

    gb7nash

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    It depends on the value of k. As stated before, y1 and y2 are solutions for -9/4 <= k < 0, and only y2 is a solution for k >= 0. Since this is the case, a) is what you're looking for.
     
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