The following quadratic equation is given:
\(x^2 - (a - 2)x - (a + 1) = 0\)
The equation is in the form:
\(x^2 + px + q = 0\)
where $p = -(a - 2) = -a + 2$ and $q = -(a + 1) = -a - 1$.
For the equation $x^2 + px + q = 0$, Vieta’s formulas relate coefficients and roots. Let the roots be $\alpha$ and $\beta$. Vieta’s formulas state:
$\alpha + \beta = -p$ and $\alpha \beta = q$
Therefore:
$\alpha + \beta = a - 2$ and $\alpha \beta = -a - 1$
We need to find $\alpha^2 + \beta^2$.
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\)
Substituting from Vieta's formulas:
\(\alpha^2 + \beta^2 = (a - 2)^2 - 2(-a - 1)\)
Simplifying:
\(\alpha^2 + \beta^2 = (a - 2)^2 + 2(a + 1)\)
Expanding:
\(\alpha^2 + \beta^2 = (a^2 - 4a + 4) + 2a + 2 = a^2 - 2a + 6\)
We want to minimize $a^2 - 2a + 6$. A quadratic of the form $ax^2 + bx + c$ is minimized at $x = -\frac{b}{2a}$.
For $a^2 - 2a + 6$, $a = 1$, $b = -2$, and $c = 6$. The value of $a$ that minimizes the expression is:
\(a = -\frac{-2}{2(1)} = 1\)
The value of $a$ that minimizes the sum of the squares of the roots is $a = 1$.