The roots of the quadratic equation (a + b - | vedclass

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(D) Given the quadratic equation $(a + b - 2c)x^2 - (2a - b - c)x + (a - 2b + c) = 0$.Let the coefficients be $A = (a + b - 2c)$,$B = -(2a - b - c)$,a

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None of these

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(D) Given the quadratic equation $(a + b - 2c)x^2 - (2a - b - c)x + (a - 2b + c) = 0$.Let the coefficients be $A = (a + b - 2c)$,$B = -(2a - b - c)$,and $C = (a - 2b + c)$.Calculate the sum of the coefficients: $A + B + C = (a + b - 2c) - (2a - b - c) + (a - 2b + c) = a + b - 2c - 2a + b + c + a - 2b + c = (a - 2a + a) + (b + b - 2b) + (-2c + c + c) = 0 + 0 + 0 = 0$.Since the sum of the coefficients is $0$,one root of the equation must be $x = 1$.Let the other root be $\alpha$. The product of the roots is given by $\frac{C}{A} = \frac{a - 2b + c}{a + b - 2c}$.Since $1 \cdot \alpha = \frac{a - 2b + c}{a + b - 2c}$,the roots are $1$ and $\frac{a - 2b + c}{a + b - 2c}$.Comparing this with the given options,none of the options $A$,$B$,or $C$ match the result. Therefore,the correct answer is $D$.

The roots of the quadratic equation $(a + b - 2c)x^2 - (2a - b - c)x + (a - 2b + c) = 0$ are

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