Question

if x = 3+2√2 find √x + 1/√x

Answer 1

x = 3 + 2√2

$\frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} }$
rationalising, we get

$\frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \\ \\ \frac{1}{x} = > \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ = > \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{ {3}^{2} - (2 \sqrt{2)} {}^{2} } \\ = > \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ = > \frac{1}{x} = 3 - 2 \sqrt{2}$
Now , value of √x =

$\sqrt{3 + 2 \sqrt{2} }$
Let,
$\sqrt{3 + 2 \sqrt{2} } = \sqrt{a} + \sqrt{b} \\ = > ( \sqrt{3 + 2 \sqrt{2} } ) {}^{2} = ( \sqrt{a} + \sqrt{b} ) {}^{2} \\ = > 3 + 2 \sqrt{2} = a + b + 2 \sqrt{ab}$
Comparing, we get

a + b = 3
and 2√(ab) = 2√2

=> √(ab )= √2

=> ab = 2

Now only one possible number can satisy this equation that is
a = 2 and b = 1
a + b = 2 + 1 = 3
ab = 2 × 1 = 2

Hence a = 2 and b = 1

Also, we know that,
$\sqrt{x} = \sqrt{3 + 2 \sqrt{2} } = \sqrt{a} + \sqrt{b} \\ = > \sqrt{x} = \sqrt{a} + \sqrt{b} \\ = > \sqrt{x} = \sqrt{2} + \sqrt{1} \\ = > \sqrt{x} = \sqrt{2} + 1$
Now we know that 1/x = 3 -2√2

Hence by following the process of √x,
1/√x = √2 - 1

Adding √x + 1/√x

= √2 + 1 + √2 - 1

=> 2√2

Hence your answer is 2√2

Hope it helps dear friend ☺️✌️