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

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

Sep 28th, 2015

Thank you for the opportunity to help you with your question!

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


Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 28th, 2015

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