Find that non-zero value of k, 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.

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Solution

We have
$k x^{2} + 1 - 2 (k - 1) x + x^{2} = 0$
This equation can be rearranged as
$(k + 1) x^{2} - 2 (k - 1) x + 1 = 0$
Here, a = k + 1, b = -2(k - 1) and c = 1
$∴ D = b^{2} - 4 a c$
$[- 2 (k - 1)^{2}] - 4 \times (k + 1) \times 1$
$= 4 [(k - 1)^{2} - (k + 1)]$
$= 4 [k^{2} - 3 k]$
$4 [k (k - 3)]$
The given equation will have equal roots, if D = 0.
$\Rightarrow 4 [k (k - 3)] = 0$
$\Rightarrow k = 0 o r k - 3 = 0$
$\Rightarrow k = 3$
Putting k = 3 in the given equation, we get
$4 x^{2} - 4 x + 1 = 0$
$\Rightarrow (2 x - 1)^{2} = 0$
$\Rightarrow 2 x - 1 = 0$
$\Rightarrow x = \frac{1}{2}$
Hence, the roots of the given equation are $\frac{1}{2}$ and $\frac{1}{2}$ .

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