Q34 of 40 Page 1

Find the 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.


Equal roots i.e. determinant is 0

⇒ b2 – 4ac = 0


Equation given: kx2 + 1 – 2(k – 1)x + x2 = 0


⇒ x2(k + 1) – 2(k – 1)x + 1 = 0


⇒ (-2(k – 1))2 – 4(k + 1)1 = 0


⇒ 4(k – 1)2 – 4k – 4 = 0


⇒ 4k2 – 8k + 4 – 4k – 4 = 0


⇒ 4k2 – 12k = 0


⇒ 4k(k – 3) = 0


⇒ k = 0 or k = 3


If k = 0, equation is x2 + 2x + 1 = 0


⇒ (x + 1)2 = 0


∴ roots will be -1 and -1


If k = 3, equation is 4x2 – 4x + 1 = 0


⇒ (2x – 1)2 = 0


∴ roots will be 1/2 and 1/2 .


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