Question

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

Answer 1

Solution:-

Given :-

$\rm \: x = 3 + 2 \sqrt{2}$

To find :-

$\rm \: {x}^{2} + \frac{1}{x {}^{2} }$

Now put the value :-

$\rm \: (3 + 2 \sqrt{2} ) {}^{2} + \frac{1}{(3 + 2 \sqrt{2} ) {}^{2} } {}^{}$

Using this identity

$\rm \: (a \: + \: b) {}^{2} = {a}^{2} + {b}^{2} + 2ab$

we get

$(3) {}^{2} + (2 \sqrt{2} ) {}^{2} + 2 \times 3 \times 2 \sqrt{2} + \frac{1}{(3) {}^{2} + (2 \sqrt{2} ) {}^{2} + 2 \times 3 \times 2 \sqrt{2} }$

$\rm \: 9 + 8 + 12 \sqrt{2} + \frac{1}{9 + 8 + 12 \sqrt{2} }$

$\rm \: 17 + 12 \sqrt{2} + \frac{1}{17 + 12 \sqrt{2} }$

Again take lcm

$\rm \: \frac{(17 + 12 \sqrt{2} ) {}^{2} + 1}{17 + 12 \sqrt{2} }$

$\rm \: \frac{(17) {}^{2} + (12 \sqrt{2}) {}^{2} + 2 \times 17 \times 12 \sqrt{2} + 1 }{17 + 12 \sqrt{2} }$

$\rm \: \frac{289 + 288 + 408 \sqrt{2} + 1 }{17 + 12 \sqrt{2} }$

$\rm \: \frac{289 + 289 + 408 \sqrt{2} }{17 + 12 \sqrt{2} }$

$\rm \: \frac{578 + 408 \sqrt{2} }{17 + 12 \sqrt{2} }$

$\rm \: \frac{34(17 + 12 \sqrt{2} )}{17 + 12 \sqrt{2} }$

$\rm \: \to \: 34$

Answer:- 34

Answer 2

$\sf{\underline{\boxed{\large{\blue{\mathsf{Question}}}}}}$

• if $\sf x = 3 + 2\sqrt 2$ , find the value of $\sf x^2 + \dfrac{1}{x}^2$

_______________________

$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• $\sf x = 3 + 2\sqrt 2$

_______________________

$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• $\sf x^2 + \dfrac{1}{x}^2$

______________________

$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

$\sf{\red{\boxed{\bold{x = 3 + 2\sqrt 2 }}}}$

• Finding the value of $\sf\dfrac{1}{x}$

$\sf\implies \dfrac{1}{3 + 2\sqrt 2}$

Rationalize the denominator

$\sf\implies \dfrac{1}{3 + 2\sqrt 2}\times\dfrac{3 - 2\sqrt 2 }{ 3 - 2 \sqrt 2}$

$\sf{\orange{\boxed{\bold{( a + b ) ( a - b ) = a^2 - b^2 }}}}$

$\sf\implies \dfrac{3 - 2\sqrt 2 }{3^2 + (2\sqrt 2)^2}$

$\sf\implies \dfrac{ 3 - 2 \sqrt 2}{9 - 8 }$

$\sf\implies \dfrac{ 3 - 2\sqrt 2}{1}$

$\sf{\red{\boxed{\bold{\dfrac{1}{x} = 3 - 2\sqrt 2 }}}}$

______________________

$\sf{\orange{\boxed{\bold{ ( x + \dfrac{1}{x})^2 = x^2+ \dfrac{1}{x^2} + 2 }}}}$

$\sf\implies ( 3 +2\sqrt 2 + 3 - 2\sqrt 2 )^2 = x^2 +\dfrac{1}{x^2} + 2$

$\sf\implies ( 3 + 3 )^2 = x^2 +\dfrac{1}{x^2} + 2$

$\sf\implies ( 6 )^2 = x^2 +\dfrac{1}{x^2} + 2$

$\sf\implies 36 = x^2 +\dfrac{1}{x^2} + 2$

$\sf\implies 36 - 2 = x^2 +\dfrac{1}{x^2}$

$\sf\implies x^2 +\dfrac{1}{x^2} = 34$

$\sf{\red{\boxed{\bold{x^2 + \dfrac{1}{x^2} = 34 }}}}$

________________________

$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

• $\sf\longrightarrow\dag x +\dfrac{1}{x^2} = 34$