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Step-by-step Solution
Step 1: Identify the sum and product of the roots
The given quadratic equation is
$$x^2 - (a - 2)x - (a + 1) = 0.$$
By comparing with the standard form
$$x^2 + px + q = 0,$$
we see that:
Sum of roots, $ \alpha + \beta = a - 2.$
Product of roots, $ \alpha \beta = -(a + 1).$
Step 2: Express the sum of the squares of roots
We use the identity
$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta.$$
Substituting the values from Step 1:
$$\alpha^2 + \beta^2 = (a - 2)^2 - 2 \bigl[-(a + 1)\bigr].$$
Step 3: Simplify the expression
Expand and simplify:
$$
(a - 2)^2 = a^2 - 4a + 4,
$$
$$
-2[-(a+1)] = 2(a + 1) = 2a + 2.
$$
Thus,
$$
\alpha^2 + \beta^2 = a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6.
$$
Step 4: Find the minimum value of the sum of squares
The expression
$$a^2 - 2a + 6$$
can be written in a completed square form:
$$
a^2 - 2a + 6 = (a - 1)^2 + 5.
$$
The term $(a - 1)^2$ is always non-negative (i.e., $(a - 1)^2 \geq 0$). Therefore, the smallest it can be is $0$, occurring when $a = 1$. At that point, the expression becomes:
$$
(a - 1)^2 + 5 \Big|_{a=1} = 0 + 5 = 5.
$$
Step 5: Determine the value of $a$ for the minimum sum of squares
The sum of the squares of the roots reaches its minimum value when $(a - 1)^2$ is zero, which occurs at $a = 1.$ Hence,
The value of $a$ that makes the sum of the squares of the roots minimum is $1.$
