Find all values of $a$ for which the sum of the roots of the equation $x^{2} - 2 a (x - 1) - 1 = 0$ is equal to the sum of the squares of its roots.

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Solution

$x^{2} - 2 a (x - 1) - 1 = 0$
$x^{2} - 2 a x + 2 a - 1 = 0$
Let roots be $α, β$
Sum of roots $= α + β = 2 a$
Product of roots $= α β = 2 a^{- 1}$
$α^{2} + β^{2} = α + β$
$(α + β)^{2} - 2 α β = α + β$
$(2 a)^{2} - 2 (2 a - 1) = 2 a$
$4 a^{2} - 4 a + 2 = 2 a$
$4 a^{2} - 6 a + 2 = 0$
$2 a^{2} - 3 a + 1 = 0$
$2 a^{2} - 2 a - a + 1 = 0$
$(2 a - 1) (a - 1) = 0$
$a = 1$ or $\frac{1}{2}$

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