The roots of the quadratic equation $(a+b-2c)x^2-(2a-b-c)x+(a-2b+c)=0$ are-
- $(a+b+c)$ and $(a-b-c)$
- $\dfrac 12$ and $a-2b+c$
- $a-2b+c$ and $\dfrac 1{(a+b-2c)}$
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None of the above.
Let f(x) = (a+b-2c)x^2 - (2a-b-c)x + (a-2b+c). Evaluating f(1) gives (a+b-2c) - (2a-b-c) + (a-2b+c) = a-a+b+b-2b-2c+c+c = 0, so x=1 is a root. The product of roots is (a-2b+c)/(a+b-2c), so the other root is (a-2b+c)/(a+b-2c). Neither option A, B, nor C matches these values.
Check the roots by evaluating the equation at x equals 1, which results in (a plus b minus 2c) minus (2a minus b minus c) plus (a minus 2b plus c), and simplifying to zero. This means one root is 1, and by the product of roots formula, the second root must be (a minus 2b plus c) divided by (a plus b minus 2c). None of the listed options correctly pair the root 1 with this second fraction, so none of the above is the correct choice.