Find the values of a and b in each of the
`(2+sqrt(3))/(2-sqrt(3))= a+ bsqrt(3)`

Answer

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

### Step 1: Rationalize the Denominator
We will multiply the numerator and the denominator by the conjugate of the denominator, which is \(2 + \sqrt{3}\).

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

### Step 2: Expand the Numerator
Now we will expand the numerator using the formula \((a+b)^2 = a^2 + 2ab + b^2\):

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

### Step 3: Expand the Denominator
Now we will simplify the denominator using the difference of squares formula \(a^2 - b^2\):

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

### Step 4: Combine the Results
Now we can combine the results from the numerator and denominator:

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

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

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

From this, we can see that:

\[ a = 7 \quad \text{and} \quad b = 4 \]

### Final Answer
Thus, the values of \(a\) and \(b\) are:

\[ a = 7, \quad b = 4 \]

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