The roots of the equation $(a + b - c) x^{2} - 2 a x + (a - b + c) = 0$ for all $a, b, c \in Q$ are

1, - 1

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1, \frac{a - b + c}{a + b - c}

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1, \frac{a + b - c}{a - b + c}

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Solution

The correct option is B $1, \frac{a - b + c}{a + b - c}$
Given:
$(a + b - c) x^{2} - 2 a x + (a - b + c) = 0$ for all $a, b, c \in Q$
Now, the sum of coefficients is:
$(a + b - c) - 2 a + (a - b + c) = 0$
Now, we know for any quadratic equation $p x^{2} + q x + r = 0$
If $p + q + r = 0 \Rightarrow 1$ is one of the roots of the equation.
And the other root is given by:
$\frac{constant term}{coefficient of x^{2}} = \frac{a - b + c}{a + b - c}$
Thus, the roots of the euqation are:
$1, \frac{a - b + c}{a + b - c}$

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The roots of the equation (a+b−c)x2−2ax+(a−b+c)=0 for all a,b,c∈Q are

The roots of the equation $(a + b - c) x^{2} - 2 a x + (a - b + c) = 0$ for all $a, b, c \in Q$ are