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(C) Let the roots of the quadratic equation $x^{2}+(3-a)x+(1-2a)=0$ be $\alpha$ and $\beta$.From the relation between roots and coefficients,we have $

Vedclass · Accepted Answer

$6$

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(C) Let the roots of the quadratic equation $x^{2}+(3-a)x+(1-2a)=0$ be $\alpha$ and $\beta$.From the relation between roots and coefficients,we have $\alpha+\beta = -(3-a) = a-3$ and $\alpha\beta = 1-2a$.The sum of the squares of the roots is given by $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2}-2\alpha\beta$.Substituting the values,we get $f(a) = (a-3)^{2} - 2(1-2a)$.Expanding this,$f(a) = a^{2} - 6a + 9 - 2 + 4a = a^{2} - 2a + 7$.To find the minimum value,we complete the square: $f(a) = (a^{2} - 2a + 1) + 6 = (a-1)^{2} + 6$.Since $(a-1)^{2} \geq 0$,the minimum value of $f(a)$ is $6$ when $a=1$.

The minimum value of the sum of the squares of the roots of $x^{2}+(3-a)x+1=2a$ is:

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