The value of a for which the sum of the squar | vedclass

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

Vedclass · Accepted Answer

$1$

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(B) Let $\alpha$ and $\beta$ be the roots of the equation $x^2-(a-2)x-(a+1)=0$. From the relation between roots and coefficients, we have $\alpha+\beta = a-2$ and $\alpha\beta = -(a+1)$. 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 $\alpha^2+\beta^2 = (a-2)^2 - 2(-(a+1)) = (a-2)^2 + 2(a+1)$. Expanding this, we get $\alpha^2+\beta^2 = a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6$. To find the minimum value, we complete the square: $a^2 - 2a + 6 = (a^2 - 2a + 1) + 5 = (a-1)^2 + 5$. Since $(a-1)^2 \geq 0$, the minimum value is $5$, which occurs when $a-1 = 0$, i.e.,$a = 1$.

The value of $a$ for which the sum of the squares of the roots of the equation $x^2-(a-2)x-(a+1)=0$ assumes the least value is

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