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Question
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.
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Solution
We have
kx2 +1−2(k−1)x+x2=0
This equation can be rearranged as
(k+1)x2 −2(k−1)x+1=0
Here, a = k + 1, b = −2(k − 1) and c = 1
∴ D = b2 − 4ac
=[−2(k−1)2]−4×(k+1)×1
=4(k−1)2−4(k+1)
=4[(k−1)2 −k−1]
=4[k2 +1−2k−k−1]
=4[k2−3k]
=4[k(k−3)]
The given equation will have equal roots, if D = 0
⇒ 4[k(k−3)] = 0
⇒ k = 0 or k − 3 = 0
⇒ k = 3
Putting k = 3 in the given equation, we get
4x2−4x+1=0
⇒(2x−1)2=0
⇒2x−1=0
`=>x=1/2`
Hence, the roots of the given equation are `1/2 " and "1/2`
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