Question

Find all possible values of a, such that x^2−2ax+2a−1=0 has exactly two distinct roots.

Answer 1

We could do this a couple different ways,

Way One.

$x^2-2ax+(2a-1)=0$

$(x-1)(x - (2a-1))=0$

That has roots x=1 and x=2a-1

We need them to be different.

$1 \ne 2a - 1$

$2 \ne 2a$

$a \ne 1$

That's the answer.

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Way Two.

$x^2-2ax+(2a-1)=0$

has two distinct roots when the discriminant is strictly positive.

$D = (-2a)^2 - 4(1)(2a - 1) = 4a^2 - 8a + 4 = 4(a-1)^2$

We need D>0.  Since squares of reals are never negative, D>0 will be true whenever a ≠ 1.

Answer: a ≠ 1