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Find k so that the roots kx^2-6x+3=0 are equal?
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(sqrt(k)x - sqrt(3))(sqrt(k)x - sqrt(3)) = 0 2(sqrt(k))(sqrt(3)) = 6 sqrt(k)sqrt(3) = 3 k = 3 here's the graph with the single root: http://bit.ly/qXZffP
k=0
which one is correct?
mine.
sqrt?
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square root.
\[(\sqrt{k}x-\sqrt{3})(\sqrt{k}x-\sqrt{3})\]
multiplied out is \[kx^2 -2\sqrt{k}\sqrt{3}x+3\] so \[2\sqrt{k}\sqrt{3} = 6\]\[\sqrt{k}\sqrt{3}=3\]\[\sqrt{k}=\sqrt{3}\]\[k=3\]
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