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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- anonymous

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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- IrishBoy123

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

- anonymous

(x-a)(x-a)
x^2-ax-ax+a^2
I don't quite understand how this relates to the problem?

- IrishBoy123

\(x^2-ax-ax+a^2 = \\ x^2 - 2ax + a^2\)
original problem:
\(x^2-2x+k=0\)
geddit?

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## More answers

- anonymous

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.

- IrishBoy123

pattern match

- IrishBoy123

|dw:1439636560854:dw|

- anonymous

All right. But how does that fit in? must i make a variable 'a' and solve for that first or something?

- anonymous

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

- IrishBoy123

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

- IrishBoy123

typo:
"k = -9 gives you the ***other*** solution"

- anonymous

Ok! That really helps! One little thing though, so does (2k)^2=4k^2 then?

- IrishBoy123

\((2k)^2=4k^2\) \(\checkmark\)

- anonymous

Thank you very very much! That has helped me a lot!

- anonymous

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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