If one root is the `n^("th")` power of the other root of the equation `x^(2) -ax + b = 0` then `b^((n)/(n+1))+b^((1)/(n+1))`=

Options

A. ab
B. `a^(n)`
C. a
D. `b^(n)`

Answer

To solve the problem, we need to find the value of \( b^{\frac{n}{n+1}} + b^{\frac{1}{n+1}} \) given that one root of the quadratic equation \( x^2 - ax + b = 0 \) is the \( n^{th} \) power of the other root.

### Step-by-Step Solution:

1. **Define the Roots**:
Let the roots of the equation be \( \alpha \) and \( \alpha^n \), where \( \alpha^n \) is the \( n^{th} \) power of \( \alpha \).

2. **Sum of Roots**:
According to Vieta's formulas, the sum of the roots is given by:
\[ \alpha + \alpha^n = a \]
This gives us our first equation.

3. **Product of Roots**:
The product of the roots is also given by Vieta's formulas:
\[ \alpha \cdot \alpha^n = b \]
Simplifying this, we get:
\[ \alpha^{n+1} = b \]
This gives us our second equation.

4. **Express \( \alpha \)**:
From the equation \( \alpha^{n+1} = b \), we can express \( \alpha \) as:
\[ \alpha = b^{\frac{1}{n+1}} \]

5. **Substitute \( \alpha \) into the Sum of Roots**:
Now, substitute \( \alpha = b^{\frac{1}{n+1}} \) into the sum of roots equation:
\[ b^{\frac{1}{n+1}} + (b^{\frac{1}{n+1}})^n = a \]
This simplifies to:
\[ b^{\frac{1}{n+1}} + b^{\frac{n}{n+1}} = a \]

6. **Rearranging the Equation**:
We need to find \( b^{\frac{n}{n+1}} + b^{\frac{1}{n+1}} \). From the equation we derived:
\[ b^{\frac{n}{n+1}} + b^{\frac{1}{n+1}} = a \]

7. **Final Expression**:
Thus, we conclude that:
\[ b^{\frac{n}{n+1}} + b^{\frac{1}{n+1}} = a \]

### Conclusion:
The value of \( b^{\frac{n}{n+1}} + b^{\frac{1}{n+1}} \) is equal to \( a \).