Find the value of $a$ such that the sum of the squares of the roots of the equation $x^{2}-(a-2) x-(a+1)=0$ is least.

Step 1: Identify coefficients and root properties

For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of the roots is given by $\alpha + \beta = -B/A$ and the product of the roots is given by $\alpha \beta = C/A$. In our given equation, $A=1$, $B=-(a-2)$, and $C=-(a+1)$.

Step 2: Express sum and product of roots

Using the formulas for the sum and product of roots, we substitute the coefficients from our equation. The sum of the roots $\alpha + \beta$ becomes $a-2$, and the product of the roots $\alpha \beta$ becomes $-(a+1)$.

Step 3: Formulate sum of squares of roots

We want to find the value of $a$ for which the sum of the squares of the roots, $\alpha^2 + \beta^2$, is least. We can express $\alpha^2 + \beta^2$ in terms of the sum and product of the roots using the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$.

Step 4: Substitute and simplify the expression

Substitute the expressions for $(\alpha + \beta)$ and $\alpha\beta$ into the formula for $\alpha^2 + \beta^2$. Expand and simplify the resulting quadratic expression in terms of $a$.

Step 5: Find the minimum value of the quadratic

The expression for the sum of the squares of the roots is a quadratic function of $a$, $f(a) = a^2 - 2a + 6$. This is a parabola opening upwards, so its minimum value occurs at the vertex. We find the vertex by setting the first derivative with respect to $a$ to zero.

Step 6: Verify minimum using second derivative

The second derivative of $f(a)$ is $2$, which is positive. This confirms that $a=1$ corresponds to a local minimum.

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