The quadratic equation kx^2+(k-3)x+1=0 has two equal real roots. Find the possible values of k.

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This is how you find roots in quadratic equation

(-b+- sqrt(b^2-4ac))/2a

if the roots are equal that means b^2-4ac=0

so

(k-3)^2-4k=0

k^2-6k+9-4k=0

k^2-10k+9=0

=> k=5+-sqrt(25-9) so k=9 or k=1

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