Please explain how to find the value for k for which this equation has one repeated root x^2-2x+k=0
Stacey Warren - Expert brainly.com
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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
I don't quite understand how this relates to the problem?
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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.
All right. But how does that fit in? must i make a variable 'a' and solve for that first or something?
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
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
\( 3x^2 +2kx-3k = 0\)
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
"k = -9 gives you the ***other*** solution"
Ok! That really helps! One little thing though, so does (2k)^2=4k^2 then?
Thank you very very much! That has helped me a lot!
solve for k.
Add -x^2 to both sides.
Add 2x to both sides.
k = −x^2+2x