A quadratic equation whose roots are `(a)/(sqrt(a) pm sqrt((a-b)))` is

Options

A. `ax^(2)-b sqrt(bx) +b^(2)=0`
B. `bx^(2)-2a sqrt(ax) +a^(2)=0`
C. `ax^(2)-abx+b=0`
D. `a^(2)x^(2)+abx+b^(2)=0`

Answer

To find the quadratic equation whose roots are given as \( \frac{a}{\sqrt{a} + \sqrt{a-b}} \) and \( \frac{a}{\sqrt{a} - \sqrt{a-b}} \), we will follow these steps:

### Step 1: Identify the Roots
Let:
- \( r_1 = \frac{a}{\sqrt{a} + \sqrt{a-b}} \)
- \( r_2 = \frac{a}{\sqrt{a} - \sqrt{a-b}} \)

### Step 2: Calculate the Sum of the Roots
The sum of the roots \( S \) is given by:
\[ S = r_1 + r_2 = \frac{a}{\sqrt{a} + \sqrt{a-b}} + \frac{a}{\sqrt{a} - \sqrt{a-b}} \]

To add these fractions, we need a common denominator:
\[ S = \frac{a(\sqrt{a} - \sqrt{a-b}) + a(\sqrt{a} + \sqrt{a-b})}{(\sqrt{a} + \sqrt{a-b})(\sqrt{a} - \sqrt{a-b})} \]
\[ = \frac{a(\sqrt{a} - \sqrt{a-b} + \sqrt{a} + \sqrt{a-b})}{a - (a-b)} = \frac{2a\sqrt{a}}{b} \]

### Step 3: Calculate the Product of the Roots
The product of the roots \( P \) is given by:
\[ P = r_1 \cdot r_2 = \left(\frac{a}{\sqrt{a} + \sqrt{a-b}}\right) \cdot \left(\frac{a}{\sqrt{a} - \sqrt{a-b}}\right) \]
\[ = \frac{a^2}{(\sqrt{a} + \sqrt{a-b})(\sqrt{a} - \sqrt{a-b})} = \frac{a^2}{a - (a-b)} = \frac{a^2}{b} \]

### Step 4: Form the Quadratic Equation
Using the sum and product of the roots, we can form the quadratic equation:
\[ x^2 - Sx + P = 0 \]
Substituting the values of \( S \) and \( P \):
\[ x^2 - \frac{2a\sqrt{a}}{b}x + \frac{a^2}{b} = 0 \]

### Step 5: Eliminate the Denominator
To eliminate the denominator \( b \), multiply the entire equation by \( b \):
\[ bx^2 - 2a\sqrt{a}x + a^2 = 0 \]

### Final Quadratic Equation
Thus, the quadratic equation whose roots are \( \frac{a}{\sqrt{a} + \sqrt{a-b}} \) and \( \frac{a}{\sqrt{a} - \sqrt{a-b}} \) is:
\[ bx^2 - 2a\sqrt{a}x + a^2 = 0 \]