Let `alpha` (a) and `beta` (a) be the roots of the equation `(root(3)(1+a)-1)x^(2)+(sqrt(1+a)-1) x+(root(6)(1+a)-1)=0, a gt -1`
Then `underset(a rarr 0^(+)) lim alpha" (a) and "underset(a rarr 0^(+)) lim beta` (a) are

Options

A. `(-5)/(2) and 1`
B. `-(1)/(2) and -1`
C. `-(7)/(2) and 2`
D. `-(9)/(2) and 3`

Answer

To solve the problem, we need to find the limits of the roots \(\alpha(a)\) and \(\beta(a)\) of the quadratic equation given by:

\[ (\sqrt[3]{1+a}-1)x^2 + (\sqrt{1+a}-1)x + (\sqrt[6]{1+a}-1) = 0 \]

as \(a\) approaches \(0^+\).

### Step 1: Identify the coefficients of the quadratic equation

The coefficients of the quadratic equation can be defined as follows:

- \(A = \sqrt[3]{1+a} - 1\)
- \(B = \sqrt{1+a} - 1\)
- \(C = \sqrt[6]{1+a} - 1\)

### Step 2: Apply the quadratic formula

The roots of the quadratic equation \(Ax^2 + Bx + C = 0\) can be found using the quadratic formula:

\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]

### Step 3: Calculate the limits as \(a \to 0^+\)

We need to evaluate:

\[ \lim_{a \to 0^+} \alpha(a) \quad \text{and} \quad \lim_{a \to 0^+} \beta(a) \]

This requires substituting \(a = 0\) into the coefficients \(A\), \(B\), and \(C\).

#### Step 3.1: Evaluate \(A\), \(B\), and \(C\) at \(a = 0\)

- \(A = \sqrt[3]{1+0} - 1 = 0\)
- \(B = \sqrt{1+0} - 1 = 0\)
- \(C = \sqrt[6]{1+0} - 1 = 0\)

At \(a = 0\), all coefficients \(A\), \(B\), and \(C\) become \(0\). This means we have an indeterminate form \(0/0\).

### Step 4: Use L'Hôpital's Rule

Since we have a \(0/0\) form, we can apply L'Hôpital's Rule. We will differentiate the numerator and the denominator with respect to \(a\).

#### Step 4.1: Differentiate \(A\), \(B\), and \(C\)

1. Differentiate \(A\):
\[ A' = \frac{d}{da}(\sqrt[3]{1+a} - 1) = \frac{1}{3(1+a)^{2/3}} \]

2. Differentiate \(B\):
\[ B' = \frac{d}{da}(\sqrt{1+a} - 1) = \frac{1}{2\sqrt{1+a}} \]

3. Differentiate \(C\):
\[ C' = \frac{d}{da}(\sqrt[6]{1+a} - 1) = \frac{1}{6(1+a)^{5/6}} \]

### Step 5: Evaluate the limits of the roots

Now we can substitute these derivatives back into the quadratic formula and evaluate the limits as \(a \to 0^+\).

#### Step 5.1: Substitute and simplify

Using the derivatives, we can find the limits of \(\alpha(a)\) and \(\beta(a)\):

\[ \lim_{a \to 0^+} \frac{-B' \pm \sqrt{(B')^2 - 4A'C'}}{2A'} \]

Substituting \(a = 0\):

- \(A' = \frac{1}{3}\)
- \(B' = \frac{1}{2}\)
- \(C' = \frac{1}{6}\)

Now substituting these values into the limit:

\[ \lim_{a \to 0^+} \frac{-\frac{1}{2} \pm \sqrt{\left(\frac{1}{2}\right)^2 - 4 \cdot \frac{1}{3} \cdot \frac{1}{6}}}{2 \cdot \frac{1}{3}} \]

Calculating the discriminant:

\[ \left(\frac{1}{2}\right)^2 - 4 \cdot \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{4} - \frac{2}{9} = \frac{9}{36} - \frac{8}{36} = \frac{1}{36} \]

Thus, we have:

\[ \sqrt{\frac{1}{36}} = \frac{1}{6} \]

Now substituting back:

\[ \lim_{a \to 0^+} \frac{-\frac{1}{2} \pm \frac{1}{6}}{\frac{2}{3}} = \frac{-\frac{1}{2} \pm \frac{1}{6}}{\frac{2}{3}} = \frac{-\frac{3}{6} \pm \frac{1}{6}}{\frac{2}{3}} = \frac{-\frac{2}{6}}{\frac{2}{3}} = -\frac{1}{2} \cdot \frac{3}{2} = -\frac{3}{4} \]

### Final Result

Thus, the limits are:

\[ \lim_{a \to 0^+} \alpha(a) = -\frac{3}{4}, \quad \lim_{a \to 0^+} \beta(a) = -\frac{3}{4} \]