question_answer

Find that non-zero value of fc, for which the quadratic equation \[k{{x}^{2}}+1-2(k-1)x+{{x}^{2}}=0\] has equal roots. Hence find the roots of the equation.

Answer:

The given equation can be written as

\[(k+1){{x}^{2}}-2(k-1)x+1=0\]

Since the equation has equal roots

\[4{{(k-1)}^{2}}-4(k+1)=0\]

\[4({{k}^{2}}+1-2k)-4(k+1)=0\]

\[4{{k}^{2}}+4-8k-4k-4=0\]

\[4{{k}^{2}}-12k=0\]

\[4k(k-3)=0\]

\[k=0,3\]

\[\therefore \] Non zero value of k is 3.

And the equation becomes,

\[4{{x}^{2}}-4x+1=0\]

\[{{(2x-1)}^{2}}=0\]

\[x=\frac{1}{2},\frac{1}{2}\] which are the required roots of the given equation.