1.

If x = 3 + 2√2, the value of \({x^2} + \frac{1}{{{x^2}}}\) is1). 362). 303). 324). 34

Answer»

Formula:

(a + B)2 = a2 + b2 + 2ab

(a – b)(a + b) = a2 – b2

Given,

x = 3 + 2√2

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

Multiplying and dividing by 3 – 2√2

$(\Rightarrow \frac{1}{x} = \frac{{3 - 2\sqrt 2 }}{{\left( {3 + 2\sqrt 2 } \right)\left( {3 - 2\sqrt 2 } \right)}}\;)$

⇒ 1/x = 3 – 2√2

$({\left( {x + \frac{1}{x}} \right)^2} = {x^2} + \frac{1}{{{x^2}}} + 2)$

$(\Rightarrow {\left( {3 + 2\sqrt 2+ 3 - \;2\sqrt 2 } \right)^2} - 2 = {x^2} + \frac{1}{{{x^2}}})$

$(\Rightarrow {x^2} + \frac{1}{{{x^2}}} = {6^2} - 2 = 34)$


Discussion

No Comment Found