Question

If X= 3+2√2,find the value of x^2 + 1/x^2
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Answer 1

Solution!!

Given:- x = 3 + 2√2

To find:- x² + 1/x²

x² = (3 + 2√2)²

= (3)² + (2√2)² + 2(3)(2√2)

= 9 + 8 + 12√2

x² = 17 + 12√2

1/x² = 1/(17 + 12√2)

= 1/(17 + 12√2) × (17 - 12√2)/(17 - 12√2)

= [1(17 - 12√2)]/[(17 + 12√2)(17 - 12√2)]

= [17 - 12√2]/[(17)² - (12√2)²]

= [17 - 12√2]/[289 - 288]

= (17 - 12√2)/1

1/x² = 17 - 12√2

x² + 1/x² = (17 + 12√2) + (17 - 12√2)

= 17 + 12√2 + 17 - 12√2

= 17 + 17 + 12√2 - 12√2

x² - 1/x² = 34

Identities used:-

→ (a + b)² = a² + b² + ab

→ (a + b)(a - b) = a² - b²

More identities:-

→ (a - b)² = a² + b² - 2ab

→ (a + b)³ = a³ + b³ + 3ab(a + b)

→ (a - b)³ = a³ - b³ - 3ab(a - b)

→ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

→ (x + a)(x + b) = x² + (a + b)x + ab

Answer 2

Given

• $\sf{x = 3 + 2 \sqrt{2} }$

To Find

• $\sf{Value~ of \: \: \bf~ {x}^{2} + \dfrac{1}{ {x}^{2} } }$

Solution

$\sf{x = 3 + 2 \sqrt{2} }$

On Squaring both sides

$\sf{~~~~~:~~~\implies {x}^{2} ={ (3 + 2 \sqrt{2} ) }^{2} }$

${ ~~~~~~~~~~~~~~~~~~\sf{Using ~Identity \: \: _{ \pink{(a+b)²=a²+b²+2ab}}}}$

$\sf{~~~~~:~~~\implies {x}^{2} ={ {3}^{2} + {(2 \sqrt{2} ) }^{2} } + 2 \times 3 \times 2 \sqrt{2} }$

$\sf{~~~~~:~~~\implies {x}^{2} = 9 + (4 \times 2 ) + 12 \sqrt{2} }$

$\sf{~~~~~:~~~\implies {x}^{2} = 9 +8+ 12 \sqrt{2} }$

$\sf{\pink{~~~~~:~~~\implies {x}^{2} = 17 + 12 \sqrt{2} } }$

$\sf~~~~~:~~~\implies \dfrac{1}{ {x}^{2} } = \dfrac{1}{17 + 12 \sqrt{2} } \times \dfrac{17 - 12 \sqrt{2} }{17 - 12 \sqrt{2} }$

${ ~~~~~~~~~~~~~~~~~~\sf{Using ~Identity \: \: _{ \pink{(a+b)(a-b)=a²-b²}}}}$

$\sf ~~~~~:~~~\implies\dfrac{1}{ {x}^{2} } = \dfrac{17 - 12 \sqrt{2} }{ {17}^{2} - {(12 \sqrt{2} } )^{2} }$

$\sf ~~~~~:~~~\implies \dfrac{1}{ {x}^{2} } = \dfrac{17 - 12 \sqrt{2} }{ 289- (144 \times 2) }$

$\sf ~~~~~:~~~\implies \dfrac{1}{ {x}^{2} } = \dfrac{17 - 12 \sqrt{2} }{ 289- 288 }$

$\sf\pink{ ~~~~~:~~~\implies \dfrac{1}{ {x}^{2} } = 17 - 12 \sqrt{2}}$

${ ~~~~~:~~~\implies \sf{x}^{2} + \dfrac{1}{ {x}^{2} } = (17 + 12 \sqrt{2} ) + (17 - 12 \sqrt{2}) }$

${~~~~~:~~~\implies \sf{x}^{2} + \dfrac{1}{ {x}^{2} } = 17 + \xcancel{ 12 \sqrt{2}} + 17 \xcancel{- 12 \sqrt{2}} }$

${ ~~~~~:~~~\implies \bf{\pink{\overbrace{\underbrace{\red{{x}^{2} + \dfrac{1}{ {x}^{2} } = 34}}}} }}$

$\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \bigstar \: \underline{\bf{\blue{More \: Useful \: Formula}}}\\ {\boxed{\begin{array}{cc}\dashrightarrow \sf(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab \\\dashrightarrow \sf(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab \\\dashrightarrow \sf(a + b)(a - b) = {a}^{2} - {b}^{2} \\\dashrightarrow \sf(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b) \\ \dashrightarrow\sf(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b) \\ \dashrightarrow\sf a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab) \\\dashrightarrow \sf a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab \\\dashrightarrow \sf{a²+b²=(a+b)²-2ab}\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$