By rationalising the denominator of each of the following : Find in each case, the value correct to two significant figures :
` (4)/(3sqrt2-2sqrt3) `

Answer

To solve the problem of rationalizing the denominator of \(\frac{4}{3\sqrt{2} - 2\sqrt{3}}\) and finding the value correct to two significant figures, we can follow these steps:

### Step 1: Identify the rationalizing factor
The denominator is \(3\sqrt{2} - 2\sqrt{3}\). To rationalize it, we multiply by the conjugate, which is \(3\sqrt{2} + 2\sqrt{3}\).

### Step 2: Multiply the numerator and denominator by the rationalizing factor
We have:
\[ \frac{4}{3\sqrt{2} - 2\sqrt{3}} \cdot \frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} = \frac{4(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} - 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3})} \]

### Step 3: Simplify the denominator
Using the difference of squares formula \( (A - B)(A + B) = A^2 - B^2 \):
- Here, \(A = 3\sqrt{2}\) and \(B = 2\sqrt{3}\).
- So, the denominator becomes:
\[ (3\sqrt{2})^2 - (2\sqrt{3})^2 = 9 \cdot 2 - 4 \cdot 3 = 18 - 12 = 6 \]

### Step 4: Expand the numerator
Now, simplify the numerator:
\[ 4(3\sqrt{2} + 2\sqrt{3}) = 12\sqrt{2} + 8\sqrt{3} \]

### Step 5: Combine the results
Now we have:
\[ \frac{12\sqrt{2} + 8\sqrt{3}}{6} \]
This can be simplified by dividing each term in the numerator by 6:
\[ = 2\sqrt{2} + \frac{4\sqrt{3}}{3} \]

### Step 6: Substitute the approximate values of \(\sqrt{2}\) and \(\sqrt{3}\)
Using the approximate values:
- \(\sqrt{2} \approx 1.414\)
- \(\sqrt{3} \approx 1.732\)

Substituting these values:
\[ 2\sqrt{2} + \frac{4\sqrt{3}}{3} \approx 2(1.414) + \frac{4(1.732)}{3} \]
Calculating each term:
\[ 2(1.414) \approx 2.828 \]
\[ \frac{4(1.732)}{3} \approx \frac{6.928}{3} \approx 2.309 \]
Adding these together:
\[ 2.828 + 2.309 \approx 5.137 \]

### Step 7: Round to two significant figures
Rounding \(5.137\) to two significant figures gives us \(5.14\).

### Final Answer
The value of \(\frac{4}{3\sqrt{2} - 2\sqrt{3}}\) after rationalizing the denominator and rounding to two significant figures is:
\[ \boxed{5.14} \]