For what value of a,the sum of the squares of | vedclass

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(D) Let the roots of the equation be $\alpha$ and $\beta$.From the given equation $x^2-(a-2)x-(a-1)=0$,we have $\alpha+\beta = a-2$ and $\alpha\beta =

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$1$

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(D) Let the roots of the equation be $\alpha$ and $\beta$.From the given equation $x^2-(a-2)x-(a-1)=0$,we have $\alpha+\beta = a-2$ and $\alpha\beta = -(a-1) = 1-a$.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 $\alpha^2+\beta^2 = (a-2)^2 - 2(1-a)$.Expanding this,we get $\alpha^2+\beta^2 = a^2-4a+4-2+2a = a^2-2a+2$.Completing the square,we get $\alpha^2+\beta^2 = (a-1)^2+1$.For the sum of squares to be the least,$(a-1)^2$ must be $0$,which occurs when $a=1$.

For what value of $a$,the sum of the squares of the roots of the equation $x^2-(a-2)x-a+1=0$ will have the least value?

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