Question

4.
If x + 1/x = 11 find the value of x^2 + 1/x^2 ?


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Answer 1

GIVEN :-

$\to \sf \: x + \dfrac{1}{x} = 11$

TO FIND :-

$\to \sf \: value \: of \: x ^{2} + \dfrac{1}{x^{2} }$

SOLUTION :-

$\to \sf \: x + \dfrac{1}{x} = 11$

☯ Squaring both sides,

$\\ \to \sf \bigg \{x + \dfrac{1}{x} \bigg \} ^{2} = 11 ^{2}$

☯ By using identity :- (a + b)² = a² + b² + 2ab.

$\\ \to \sf \: (x )^{2} + \left(\dfrac{1}{x} \right) ^{2} + 2 \times \cancel{x}\times \dfrac{1}{ \cancel{x} } = 121 \\ \\ \\ \to \sf \: x ^{2} + \dfrac{1}{x ^{2} } + 2 = 121 \\ \\ \\ \to \sf \: x ^{2} + \dfrac{1}{x ^{2} } = 121 - 2\\ \\ \\ \to \boxed{ \red{ \sf \: x ^{2} + \dfrac{1}{x ^{2} } = 119}}$

✒ Hence x² + 1/x² = 119.

ADDITIONAL INFORMATION :-

☢ Some Algebraic Identities :-

➳ ( x + y )² = x² + 2xy + y²

➳ ( x - y )² = x² - 2xy + y²

➳ ( x - y ) ( x -y ) = ( x - y )²

➳ ( x + y ) ( x + y ) = ( x + y )²

➳ x² - y² = ( x + y ) ( x - y )

➳ ( x + y + z )² = x² + y² + z² + 2xy + 2yz + 2zx.

➳ ( x + a ) ( x + b ) = x² + ( a + b)x + ab

Answer 2

$\sf \: x \: \frac{1}{x} = 11 \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} + \frac{1}{x} = ?$

$\sf \: x + \frac{1}{x} = 1$

$\color{blue} \bigstar \displaystyle \tt \: \sf \color{red}Squaring \: \: both \: \: sides$

$\sf \: \bigg(x + \frac{1}{x} { \bigg)}^{2} = { \bigg(11 \bigg)}^{2}$

$\sf \bigg(a + b { \bigg)}^{2} = {a}^{2} + 2ab + {b}^{2}$

$\color{red} \sf \: a = x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: b = \frac{1}{x}$

$\sf \: {x}^{2} + 2 \times x \times \frac{1}{x} + \frac{1}{ {x}^{2} } = 121$


$\sf {x}^{2} + 2 + \frac{1}{ {x}^{2} } = 121$

$\underline { \boxed{ \sf{x}^{2} + \frac{1}{ {x}^{2} } = 121 - 2 = \color{red} 119}}$