Question

if x=3+2√2 , find the value of x²+1/x²​

Answer 1

Given:-

→ x = 3+2√2

To find:-

→ Value of (x²+1/x²)

Solution:-

We know that:-

=> (a+b)² = a² + b² + 2ab

=> (x+1/x)² = x²+1/x²+2×x×1/x

=> (x+1/x)² = x²+1/x²+2

=> x²+1/x² = (x+1/x)²-2

Firstly, let's calculate the value of 1/x

=> 1/x = 1/(3+2√2)

On rationalizing 1/(3+2√2), we get:-

=> 1/(3+2√2)×(3-2√2)/(3-2√2)

=> (3-2√2)/(3)²-(2√2)²

=> 3-2√2/9-8

=> 3-2√2

Hence, value of 1/x is 3-2√2

Now:-

=> x²+1/x² = (3+2√2+3-2√2)²-2

=> x²+1/x² = (6)²-2

=> x²+1/x² = 36-2

=> x²+1/x² = 34

Thus, value of (x²+1/x²) is 34.

Answer 2

$\huge \bold \color{green} \mid{ \underline{ \underline \red{question: }}} \mid$

$\bold{if \: x = 3 + 2 \sqrt{2} } \\ \bold{ then \: find \: value \: of \: {x}^{2} + \frac{1}{{x}^{2} } }$

$\huge \bold \color{green} \mid{ \underline{ \underline \orange{answer \bf : }}} \mid$

$\bold{\frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } } \\ \\ on \: rationalizing \: we \: get \\ \\ \bold{ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } } \\ \\ \bold{ \implies \frac{1}{x } = \frac{3 - 2 \sqrt{2} }{9 - 8} } \\ \\ \bold { \implies \: \frac{1}{x} = 3 - 2 \sqrt{2} }$

now..

$\bold{x + \frac{1}{x} =( 3 + 2 \sqrt{2} ) + (3 - 2 \sqrt{2} })$

$\bold{ \implies x + \frac{1}{x} = 3 + 3 + 2 \sqrt{2} - 2 \sqrt{2} } \\ \\ \bold{ \implies \: x + \frac{1}{x} = 3 + 3 = 6}$

on squaring both sides..

$\bold{( {x + \frac{1}{x} )}^{2} = {6}^{2} }\\ \\ \bold{ \implies{x}^{2} + \frac{1}{ {x}^{2} } + 2 = 36} \\ \\ \bold{ \implies {x}^{2} + \frac{1}{ {x}^{2} } = 34}$

so the value of x²+1/x² is 34