Let `alpha(a)` and `beta(a)` be the roots of the equation `((1+a)^(1/3)-1)x^2 +((1+a)^(1/2)-1)x+((1+a)^(1/6)-1)=0` where `agt-1` then `lim_(a->0^+)alpha(a)` and `lim_(a->0^+)beta(a)`

Options

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

Explanation

To solve the problem, we need to find the limits of the roots of the given quadratic equation as \( a \) approaches \( 0^+ \). Let's break down the solution step by step.

### Step 1: Write the given equation
The equation is given as:
\[ ((1+a)^{1/3} - 1)x^2 + ((1+a)^{1/2} - 1)x + ((1+a)^{1/6} - 1) = 0 \]

### Step 2: Expand the terms using Taylor series
Since \( a \) is approaching \( 0^+ \), we can use the Taylor series expansion for \( (1 + a)^n \) around \( a = 0 \):
- For \( n = \frac{1}{3} \):
\[ (1 + a)^{1/3} \approx 1 + \frac{1}{3}a \]
- For \( n = \frac{1}{2} \):
\[ (1 + a)^{1/2} \approx 1 + \frac{1}{2}a \]
- For \( n = \frac{1}{6} \):
\[ (1 + a)^{1/6} \approx 1 + \frac{1}{6}a \]

### Step 3: Substitute the expansions into the equation
Substituting these expansions into the original equation:
\[ \left(\frac{1}{3}a\right)x^2 + \left(\frac{1}{2}a\right)x + \left(\frac{1}{6}a\right) = 0 \]

### Step 4: Factor out \( a \)
Factoring out \( a \) from the equation gives:
\[ a\left(\frac{1}{3}x^2 + \frac{1}{2}x + \frac{1}{6}\right) = 0 \]
Since \( a \neq 0 \) (as \( a \to 0^+ \)), we can ignore \( a \) and focus on the quadratic:
\[ \frac{1}{3}x^2 + \frac{1}{2}x + \frac{1}{6} = 0 \]

### Step 5: Multiply through by 6 to eliminate fractions
Multiplying the entire equation by 6 to simplify:
\[ 2x^2 + 3x + 1 = 0 \]

### Step 6: Factor the quadratic equation
Factoring the quadratic:
\[ (2x + 1)(x + 1) = 0 \]
This gives us the roots:
\[ x = -\frac{1}{2}, \quad x = -1 \]

### Step 7: Identify the roots
Let \( \alpha(a) = -\frac{1}{2} \) and \( \beta(a) = -1 \).

### Step 8: Calculate the limits as \( a \to 0^+ \)
Now we can find the limits:
\[ \lim_{a \to 0^+} \alpha(a) = -\frac{1}{2}, \quad \lim_{a \to 0^+} \beta(a) = -1 \]

### Final Answer
Thus, the limits are:
\[ \lim_{a \to 0^+} \alpha(a) = -\frac{1}{2}, \quad \lim_{a \to 0^+} \beta(a) = -1 \]

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