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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^22x+k=0
anonymous
 one year ago
Please explain how to find the value for k for which this equation has one repeated root x^22x+k=0

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1why no start with seeing what it looks like to have a repeated root, say a? so multiply out \((xa)^2\) and pattern match

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(xa)(xa) x^2axax+a^2 I don't quite understand how this relates to the problem?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\(x^2axax+a^2 = \\ x^2  2ax + a^2\) original problem: \(x^22x+k=0\) geddit?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, 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.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439636560854:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0All right. But how does that fit in? must i make a variable 'a' and solve for that first or something?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the simple ones like x^22x+k I can figure it out logically by using b^24ac=0. Then i get 44*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 +2kx3k

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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: \( 3x^2 +2kx3k = 0\) 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

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1typo: "k = 9 gives you the ***other*** solution"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok! That really helps! One little thing though, so does (2k)^2=4k^2 then?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\((2k)^2=4k^2\) \(\checkmark\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you very very much! That has helped me a lot!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0solve 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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