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anonymous

  • one year ago

Please explain how to find the value for k for which this equation has one repeated root x^2-2x+k=0

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  1. IrishBoy123
    • one year ago
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    why no start with seeing what it looks like to have a repeated root, say a? so multiply out \((x-a)^2\) and pattern match

  2. anonymous
    • one year ago
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    (x-a)(x-a) x^2-ax-ax+a^2 I don't quite understand how this relates to the problem?

  3. IrishBoy123
    • one year ago
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    \(x^2-ax-ax+a^2 = \\ x^2 - 2ax + a^2\) original problem: \(x^2-2x+k=0\) geddit?

  4. anonymous
    • one year ago
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    Sorry, no I really don't. The two equations don't look similar to me because the original equation does not have two squared variables and the middle term doesn't have two variables.

  5. IrishBoy123
    • one year ago
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    pattern match

  6. IrishBoy123
    • one year ago
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    |dw:1439636560854:dw|

  7. anonymous
    • one year ago
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    All right. But how does that fit in? must i make a variable 'a' and solve for that first or something?

  8. anonymous
    • one year ago
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    For the simple ones like x^2-2x+k I can figure it out logically by using b^2-4ac=0. Then i get 4-4*1*k =0 and can easily go from there however I don't think this works for the more complicated ones unless I am doing it wrong? The most complicated one I have to do in this exercise is 3x^2 +2kx-3k

  9. IrishBoy123
    • one year ago
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    yes, you solve ! \(-2a = -2 \implies a = 1\) \(k = a^2 \implies k = ...\) but if you're happier with the \(b^2 - 4ac = 0\) approach, which is fine, do it for: \( 3x^2 +2kx-3k = 0\) repeated root \( \implies (2k)^2 - 4(3)(-3k) = 0 \implies k(4k+36) = 0\) k = 0 means \(3x^2 = 0\) k = -9 gives you the solution

  10. IrishBoy123
    • one year ago
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    typo: "k = -9 gives you the ***other*** solution"

  11. anonymous
    • one year ago
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    Ok! That really helps! One little thing though, so does (2k)^2=4k^2 then?

  12. IrishBoy123
    • one year ago
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    \((2k)^2=4k^2\) \(\checkmark\)

  13. anonymous
    • one year ago
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    Thank you very very much! That has helped me a lot!

  14. anonymous
    • one year ago
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    solve for k. x^2−2x+k=0 Add -x^2 to both sides. x^2+k−2x+−x2=0+−x2 k−2x=−x^2 Add 2x to both sides. k−2x+2x=−x^2+2x k = −x^2+2x

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