Step-by-step Solution:

1. Analyze the given equation: The quadratic equation is: $(\sqrt[3]{1 + a} - 1){x^2} + (\sqrt {1 + a} - 1)x + (\sqrt[6]{1 + a} - 1) = 0$ Let the coefficients be $A(a)$, $B(a)$, and $C(a)$: $A(a) = \sqrt[3]{1 + a} - 1$ $B(a) = \sqrt{1 + a} - 1$ $C(a) = \sqrt[6]{1 + a} - 1$ As $a \to 0^+$, all three coefficients $A(a)$, $B(a)$, and $C(a)$ approach 0. This indicates that the quadratic equation degenerates.

2. Use a substitution to simplify the coefficients: To handle the different roots (cube root, square root, sixth root), let's make a substitution. Let $1+a = y^6$. The least common multiple of the denominators of the powers (3, 2, 6) is 6. As $a \to 0^+$, we have $1+a \to 1^+$, which implies $y^6 \to 1^+$, so $y \to 1^+$.

   Now, we can express the coefficients in terms of $y$: $A = \sqrt[3]{y^6} - 1 = y^2 - 1$ $B = \sqrt{y^6} - 1 = y^3 - 1$ $C = \sqrt[6]{y^6} - 1 = y - 1$

3. Rewrite and simplify the equation: Substituting these new expressions for the coefficients into the original equation, we get: $(y^2 - 1)x^2 + (y^3 - 1)x + (y - 1) = 0$ We can factor each coefficient using the difference of powers formulas: $(y-1)(y+1)x^2 + (y-1)(y^2+y+1)x + (y-1) = 0$ Since we are considering the limit as $y \to 1^+$, $y$ is not equal to 1, so $y-1 \neq 0$. We can divide the entire equation by the common factor $(y-1)$: $(y+1)x^2 + (y^2+y+1)x + 1 = 0$

4. Find the limiting equation: The roots of this equation in $x$, which we denote as $\alpha(a)$ and $\beta(a)$, depend on $y$ (and hence on $a$). To find the limits of these roots as $a \to 0^+$, we can take the limit of the simplified equation as $y \to 1^+$. $\lim_{y \to 1^+} \left[ (y+1)x^2 + (y^2+y+1)x + 1 \right] = 0$ Taking the limit of each coefficient:

   • $\lim_{y \to 1^+} (y+1) = 1+1 = 2$
   • $\lim_{y \to 1^+} (y^2+y+1) = 1^2+1+1 = 3$
   • The constant term is 1.

   The equation for the limits of the roots becomes: $2x^2 + 3x + 1 = 0$

5. Solve the limiting quadratic equation: We can solve this simple quadratic equation by factoring: $2x^2 + 2x + x + 1 = 0$ $2x(x+1) + 1(x+1) = 0$ $(2x+1)(x+1) = 0$ The roots are $x = -\frac{1}{2}$ and $x = -1$.

6. Conclusion: The limits of the roots of the original equation as $a \to 0^+$ are the roots of the limiting equation. Thus, $\{ \lim_{a \to 0^+} \alpha(a), \lim_{a \to 0^+} \beta(a) \} = \left\{ -1, -\frac{1}{2} \right\}$ The question asks for the values of these limits. We check the given options: A: $- {5 \over 2}$ B: $- {1 \over 2}$ C: $- {7 \over 2}$ D: $- {9 \over 2}$

   One of our calculated limits, $-1/2$, is present in the options. Since this is a single-choice question, we select the option that matches one of the possible values.

   Therefore, the correct option is B.