If `x=7-4sqrt3` then find the value of `sqrtx+1/sqrtx`

Explanation

To find the value of \(\sqrt{x} + \frac{1}{\sqrt{x}}\) given \(x = 7 - 4\sqrt{3}\), we can follow these steps:

### Step 1: Calculate \( \sqrt{x} + \frac{1}{\sqrt{x}} \)

We know that:
\[ \sqrt{x} + \frac{1}{\sqrt{x}} = a + b \]
where \(a = \sqrt{x}\) and \(b = \frac{1}{\sqrt{x}}\).

### Step 2: Square the expression

To simplify the calculation, we can square the expression:
\[ \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = x + \frac{1}{x} + 2 \]

### Step 3: Find \(x + \frac{1}{x}\)

We need to calculate \(x + \frac{1}{x}\). We already have \(x = 7 - 4\sqrt{3}\). Now, we will find \(\frac{1}{x}\).

### Step 4: Calculate \(\frac{1}{x}\)

To find \(\frac{1}{x}\), we rationalize:
\[ \frac{1}{x} = \frac{1}{7 - 4\sqrt{3}} \cdot \frac{7 + 4\sqrt{3}}{7 + 4\sqrt{3}} = \frac{7 + 4\sqrt{3}}{(7 - 4\sqrt{3})(7 + 4\sqrt{3})} \]
Calculating the denominator:
\[ (7 - 4\sqrt{3})(7 + 4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 48 = 1 \]
Thus:
\[ \frac{1}{x} = 7 + 4\sqrt{3} \]

### Step 5: Add \(x\) and \(\frac{1}{x}\)

Now we can find \(x + \frac{1}{x}\):
\[ x + \frac{1}{x} = (7 - 4\sqrt{3}) + (7 + 4\sqrt{3}) = 7 + 7 = 14 \]

### Step 6: Substitute back into the squared expression

Now substitute back into the squared expression:
\[ \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = 14 + 2 = 16 \]

### Step 7: Take the square root

Finally, take the square root to find \(\sqrt{x} + \frac{1}{\sqrt{x}}\):
\[ \sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{16} = 4 \]

### Final Answer

Thus, the value of \(\sqrt{x} + \frac{1}{\sqrt{x}}\) is \(4\).

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