Find the values of the integers a and b respectively, for which the solution of the equation `xsqrt(24)=xsqrt(3)+sqrt(6)` is `(a+sqrt(b))/(7)` .

Options

A. `4,2`
B. `2,6`
C. `3,2`
D. `9,5`

Answer

To solve the equation \( x \sqrt{24} = x \sqrt{3} + \sqrt{6} \) and find the values of integers \( a \) and \( b \) such that the solution can be expressed as \( \frac{a + \sqrt{b}}{7} \), we will follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ x \sqrt{24} = x \sqrt{3} + \sqrt{6} \]
Rearranging gives:
\[ x \sqrt{24} - x \sqrt{3} = \sqrt{6} \]
Factoring out \( x \):
\[ x (\sqrt{24} - \sqrt{3}) = \sqrt{6} \]

### Step 2: Solving for \( x \)
Now, isolate \( x \):
\[ x = \frac{\sqrt{6}}{\sqrt{24} - \sqrt{3}} \]

### Step 3: Simplifying the Denominator
Next, simplify \( \sqrt{24} \):
\[ \sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6} \]
Substituting this back into the equation gives:
\[ x = \frac{\sqrt{6}}{2\sqrt{6} - \sqrt{3}} \]

### Step 4: Rationalizing the Denominator
To rationalize the denominator, multiply the numerator and denominator by the conjugate \( 2\sqrt{6} + \sqrt{3} \):
\[ x = \frac{\sqrt{6}(2\sqrt{6} + \sqrt{3})}{(2\sqrt{6} - \sqrt{3})(2\sqrt{6} + \sqrt{3})} \]

### Step 5: Calculating the Denominator
Using the difference of squares:
\[ (2\sqrt{6})^2 - (\sqrt{3})^2 = 24 - 3 = 21 \]
Thus, the denominator simplifies to 21.

### Step 6: Calculating the Numerator
Now, calculate the numerator:
\[ \sqrt{6}(2\sqrt{6} + \sqrt{3}) = 2 \cdot 6 + \sqrt{18} = 12 + 3\sqrt{2} \]
So, we have:
\[ x = \frac{12 + 3\sqrt{2}}{21} \]

### Step 7: Expressing in the Required Form
Now, we can express \( x \) as:
\[ x = \frac{12}{21} + \frac{3\sqrt{2}}{21} = \frac{4}{7} + \frac{\sqrt{2}}{7} \]
This can be rewritten as:
\[ x = \frac{4 + \sqrt{2}}{7} \]

### Step 8: Identifying \( a \) and \( b \)
From the expression \( \frac{a + \sqrt{b}}{7} \), we can identify:
- \( a = 4 \)
- \( b = 2 \)

### Final Answer
Thus, the values of the integers \( a \) and \( b \) are:
\[ \boxed{4} \quad \text{and} \quad \boxed{2} \]