If one root of the quadratic equation `px^2 +qx + r = 0 (p != 0)` is a surd `sqrta/(sqrta+sqrt(a-b),` where `p, q, r; a, b` are all rationals then the other root is -

Options

A. `(sqrt(a))/(sqrt(a)- sqrt(a)-b)`
B. `(sqrt(a)-sqrt(a)-b)/(sqrt(b))`
C. `a+ sqrt(a(a-b))/(b)`
D. `(a+ sqrt(a(a-b)))/(b)`

Explanation

To find the other root of the quadratic equation \( px^2 + qx + r = 0 \) given that one root is \( \frac{\sqrt{a}}{\sqrt{a} + \sqrt{a-b}} \), we will follow these steps:

### Step 1: Identify the given root
Let \( \alpha = \frac{\sqrt{a}}{\sqrt{a} + \sqrt{a-b}} \).

### Step 2: Rationalize the given root
To rationalize \( \alpha \), we multiply the numerator and denominator by \( \sqrt{a} - \sqrt{a-b} \):
\[ \alpha = \frac{\sqrt{a}(\sqrt{a} - \sqrt{a-b})}{(\sqrt{a} + \sqrt{a-b})(\sqrt{a} - \sqrt{a-b})} \]
Using the difference of squares in the denominator:
\[ = \frac{\sqrt{a}(\sqrt{a} - \sqrt{a-b})}{a - (a-b)} = \frac{\sqrt{a}(\sqrt{a} - \sqrt{a-b})}{b} \]

### Step 3: Simplify the numerator
Now, simplify the numerator:
\[ = \frac{\sqrt{a} \cdot \sqrt{a} - \sqrt{a} \cdot \sqrt{a-b}}{b} = \frac{a - \sqrt{a(a-b)}}{b} \]

### Step 4: Find the other root using Vieta's formulas
According to Vieta's formulas, if \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( px^2 + qx + r = 0 \), then:
\[ \alpha + \beta = -\frac{q}{p} \quad \text{and} \quad \alpha \beta = \frac{r}{p} \]
We can express \( \beta \) as:
\[ \beta = -\frac{q}{p} - \alpha \]

### Step 5: Substitute the value of \( \alpha \)
Now substitute \( \alpha \):
\[ \beta = -\frac{q}{p} - \frac{a - \sqrt{a(a-b)}}{b} \]

### Step 6: Combine the terms
To combine the terms, we find a common denominator:
\[ \beta = -\frac{q}{p} - \frac{(a - \sqrt{a(a-b)})p}{bp} \]
This gives us:
\[ \beta = -\frac{q + (a - \sqrt{a(a-b)})p}{bp} \]

### Final Result
Thus, the other root \( \beta \) is:
\[ \beta = -\frac{q + (a - \sqrt{a(a-b)})p}{bp} \]