Question

If x + 1/x = 2, find x⁴ + 1/x⁴

Answer 1

$\red{\underline{{ \bf Question }}}$

• $\sf If \: x + \dfrac{1}{x} = 2, find \: x^{4} + \dfrac{1}{x^{4} }$

$\underline {\underline{{\purple{ \sf Given }}}}$

• $\bf x + \dfrac{1}{x} = 2$

$\underline {\underline{{ \purple{\sf To \: Find }}}}$

• $\bf x^{4} + \dfrac{1}{x^{4} }$

$\red{\underline{{ \bf Solution }}}$

Squaring both sides

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

$\dashrightarrow \sf {x}^{2} + 2\times \cancel{ x }\times \dfrac{1}{ \cancel{x}} + \dfrac{1}{ {x}^{2} } = 4$

$\dashrightarrow \sf {x}^{2} + \dfrac{1}{ {x}^{2} } = 4 -2$

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

$\blue{\sf We \: got \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 2}$

Squaring both sides

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

$\dashrightarrow \sf {x}^{4} +2 \times \cancel{{x}^{2}} \times \dfrac{1}{ \cancel{{x}^{2} }} + \dfrac{1}{ {x}^{4} } = 4$

$\dashrightarrow \sf {x}^{4} + \dfrac{1}{ {x}^{4} } = 4-2$

$\dashrightarrow \sf {x}^{4} + \dfrac{1}{ {x}^{4} } = 2$

$\bf Thus, \: x^{4} + \dfrac{1}{ x^{4}}= 2$

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Answer 2

$\Large{\underline{\sf{ Given}} : - } \\ \\ \sf \to x + \dfrac{1}{x} = 2 \\ \\ \Large{\underline{\rm{ Required \: to \: find }} : - } \\ \\ \bullet \to \: \tt value \: of \: x^4 + \dfrac{1}{x^4 } \\ \\ \Large{\underline{\mathscr{ F} \sf { ormula \: used }} : -} \\ \\ \boxed{\tt{ x = \dfrac{ - b \pm \sqrt{ b^2 - 4ac }}{ 2a } }} \\ \\ \Large{\underline{\mathrm{ S} \tt{ olution } } : -}$

$\to \sf x + \dfrac{1}{x} = 2 \\ \\ \to \sf x + \dfrac{1}{x } - 2 = 0 \\ \\ \rm Taking \: LCM \\ \\ \to \sf \dfrac{ x^2 + 1 - 2x }{x } = 2 \\ \\ \textrm{ By cross multiplication on both sides } \\ \\ \to \sf x^2 + 1 - 2x = 0 \times 2 \\ \\ \to \sf x^2 - 2x + 1 = 0 \\ \\ \texttt{ The standard form of a quadratic equation is \pink{ \tt ax^2 + bx + c = 0 }} \\ \\ \sf Compare \ the \: standard \ form \ a \ quadratic \: equation \: with \: the \: given \: quadratic \: equation \\ \\ \sf Here , \\ \rm \bullet a = 1 \\ \rm \bullet b = - 2 \\ \bullet \rm c = 1 \\ \\ \sf Using \: the \: Quadratic \: formula ; \\ \\ \sf{ x = \dfrac{ - ( - 2 ) \pm \sqrt{ ( - 2 )^2 - 4 ( 1 )( 1 )}}{ 2 ( 1 )}}$

$\sf x = \dfrac{ 2 \pm \sqrt{4 - 4 ( 1 ) } }{ 2 } \\ \\ \sf x = \dfrac{ 2 \pm \sqrt{4 - 4 } }{2 } \\ \\ \sf x = \dfrac{2 \pm \sqrt{0} }{2} \\ \\ \sf x_1 = \dfrac{2 + 0}{2} \quad , \quad x_2 = \dfrac{2 - 0}{2} \\ \\ \sf x_1 = \dfrac{2}{2} \quad , \quad x_2 = \dfrac{2}{2} \\ \\ \red{ \sf x_1 = 1 \quad \& \quad x_2 = 1 }$

$\tt Now, \: let's \: find \: out \: the \: value \: of \: x^4 +\dfrac{ 1}{ x^4 } \\ \\ \sf \to x_1 = 1 \\ \\ \to \sf { x_1}^4 = ( 1 )^4 = 1 \\ \\ \sf value \: of \: {x_1}^4 + \dfrac{1}{ {x_1}^4} \\ \\ \to \sf 1 + \dfrac{1}{1} \\ \\ \to \sf \dfrac{ 1 + 1 }{ 1 } \\ \\ \to \sf \dfrac{2}{1} \\ \\ \to \sf 2$

$\sf Similarly, \\ \\ \to \sf x_2 = 1 \\ \\ \to \sf {x_2}^4 = (1 )^4 = 1 \\ \\ \sf value \: of \: {x_2}^4 + \dfrac{1}{ {x_2}^4} \\ \\ \to \sf 1 + \dfrac{1}{1} \\ \\ \to \sf \dfrac{ 1 + 1 }{ 1 } \\ \\ \to \sf \dfrac{2}{1} \\ \\ \to \sf 2$

Since, the both values are equal . so, we can conclude that .

$\sf Hence, \\ \\ \orange{ \implies} \sf \blue{ Value \; of \; x^4 + \dfrac{1}{x^4 } = } \red{ 2 }$