Q12 of 47 Page 1

Find that non-zero value of k, for which the quadratic equation kx2 + 1 – 2(k-1)x + x2 = 0 has equal roots. Hence find the roots of the equation.

(k + 1)x2-2(k-1)x + 1 = 0

This is a quadratic equation of form ax2 + bx + c = 0


Where a = k + 1, b = -2(k-1), c = 1


For equal roots, D = 0


D = b2-4ac = 0


Substituting the values of a, b and c


(-2(k-1))2-4(k + 1)(1) = 0


⇒ 4(k2 + 1-2k)-4k-4 = 0


⇒ 4k2-12k = 0


⇒ 4k(k-3) = 0


There are two possible values of k, k = 0 and k = 3


We neglect k = 0 as on putting k = 0; the equation does not remain quadratic.


∴ k = 3 is our required value of k.


Now,



As we know roots are equal,


Therefore, each root =


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