We could do this a couple different ways,
Way One.
[tex]x^2-2ax+(2a-1)=0[/tex]
[tex](x-1)(x - (2a-1))=0[/tex]
That has roots x=1 and x=2a-1
We need them to be different.
[tex]1 \ne 2a - 1[/tex]
[tex]2 \ne 2a[/tex]
[tex]a \ne 1[/tex]
That's the answer.
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Way Two.
[tex]x^2-2ax+(2a-1)=0[/tex]
has two distinct roots when the discriminant is strictly positive.
[tex]D = (-2a)^2 - 4(1)(2a - 1) = 4a^2 - 8a + 4 = 4(a-1)^2[/tex]
We need D>0. Since squares of reals are never negative, D>0 will be true whenever a ≠ 1.
Answer: a ≠ 1
