If a and b are two rational numbers and `(2+sqrt(3))/(2-sqrt(3))=a+bsqrt(3)` , what is the value of b ?

Answer

To solve the equation \(\frac{2+\sqrt{3}}{2-\sqrt{3}} = a + b\sqrt{3}\) and find the value of \(b\), we will follow these steps:

### Step 1: Rationalize the denominator
We start with the expression \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\). To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is \(2+\sqrt{3}\):

\[ \frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})} \]

### Step 2: Simplify the denominator
Using the difference of squares identity \(a^2 - b^2\), we can simplify the denominator:

\[ (2-\sqrt{3})(2+\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \]

### Step 3: Simplify the numerator
Now, we simplify the numerator:

\[ (2+\sqrt{3})(2+\sqrt{3}) = 2^2 + 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \]

### Step 4: Combine the results
Putting it all together, we have:

\[ \frac{2+\sqrt{3}}{2-\sqrt{3}} = \frac{7 + 4\sqrt{3}}{1} = 7 + 4\sqrt{3} \]

### Step 5: Compare with \(a + b\sqrt{3}\)
Now we can compare this with the expression \(a + b\sqrt{3}\):

\[ 7 + 4\sqrt{3} = a + b\sqrt{3} \]

From this comparison, we can see that:

- \(a = 7\)
- \(b = 4\)

### Conclusion
Thus, the value of \(b\) is:

\[ \boxed{4} \]