Question

If x+1/x=2 then x^10+ 1/x^10+5=?

Answer 1

Answer

• $\tt x^{10} +\dfrac{1}{x^{10}}+5=7$

Given

• $\tt x+\dfrac{1}{x}=2$

To Find

• $\tt x^{10}+\dfrac{1}{x^{10}}+5$

Solution

$\boxed{\begin{minipage}{6cm} \bold{\tt{ \bold{x+\dfrac{1}{x}=2}}\\\\\bold{Squaring\ on\ both\ sides\ , we\ get\:,}\\\\\bold{\implies (x+\dfrac{1}{x})^2=2^2}\\\\\bold{\implies x^2+\dfrac{1}{x^2}+2.x.\dfrac{1}{x}=4}\\\\\bold{\implies x^2+\dfrac{1}{x^2}+2=4}\\\\\bold{\implies x^2+\dfrac{1}{x^2}=2...(1) }} \end{minipage}}$

$\boxed{\begin{minipage}{6cm} \tt { \tt{x+\dfrac{1}{x}=2}\\\\ \tt {Cubing\ on\ both\ sides\ , we\ get\ ,}\\\\\implies \tt{(x+\dfrac{1}{x})^3=2^3}\\\\\implies \tt{x^3+\dfrac{1}{x^3}+3.x.\dfrac{1}{x}(x+\dfrac{1}{x})=8}\\\\\implies \tt{x^3+\dfrac{1}{x^3}+6=8}\\\\\implies \tt{x^3+\dfrac{1}{x^3}=2...(2)} } \end{minipage}}$

Multiply (1) & (2) , we get ,

$\tt x^5 +\dfrac{1}{x}+x+\dfrac{1}{x^5}=2 \times 2\\\\\implies\tt (x^5+\dfrac{1}{x^5})=4-(x+\dfrac{1}{x})\\\\\implies \tt x^5+\dfrac{1}{x^5}=2$

Squaring on both sides , we get ,

$\implies \tt x^{10}+\dfrac{1}{x^{10}}+2.x^5.\dfrac{1}{x^5}=4\\\\\implies \tt x^{10}+\dfrac{1}{x^{10}}=2\\\\\implies \tt x^{10}+\dfrac{1}{x^{10}}=7-5\\\\\implies \bf x^{10}+\dfrac{1}{x^{10}}+5=7$

Answer 2

Given

$\huge{ \bf x + \dfrac{1}{x} = 2 }$

Squaring Both Sides

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

Identity Used :- (a+b)² = a² + 2ab + b²

Case 1

$\implies \bf {x}^{2} + 2 \times \cancel{x} \times \dfrac{1}{ \cancel{x}} + \dfrac{1}{ {x}^{2} } = 4$

$\implies \bf{x}^{2} + 2 + \dfrac{1}{ {x}^{2} } = 4$

$\implies \bf x^{2} + \dfrac{1}{ {x}^{2} } = 4 - 2$

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

Now, Cubing the result

Case 2

$\huge ({{ \bf x + \dfrac{1}{x} })^{3} = {2}^{3} }$

$\bf{x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x} (x + \dfrac{1}{x} ) = 8$

Identity Used :- (a+b)³ = a³ + b³ + 3ab(a+b)

$\implies \bf {x}^{3} + \dfrac{1}{ {x}^{3} } + 3(x + \dfrac{1}{x} ) = 8$

Value of $\bf x + \dfrac{1}{x}$ is given 2

$\implies \bf {x}^{3} + \dfrac{1}{ {x}^{3} } + 3(2) = 8$

$\implies \bf {x}^{3} + \dfrac{1}{ {x}^{3} } + 6 = 8$

$\implies \bf {x}^{3} + \dfrac{1}{ {x}^{3} } = 8 - 6$

$\implies \bf {x}^{3} + \dfrac{1}{ {x}^{3} } = 2$

If we Multiply Case 1 & Case 2, we get $\bf ( x + \dfrac{1}{x} ) ^{5}$

$\bf \longrightarrow ( {x}^{2} + \dfrac{1}{ {x}^{2} } )( {x}^{3} + \frac{1}{ {x}^{3} } ) = (2)(2)$

$\bf \longrightarrow {x}^{5} + \dfrac{1}{x} + x + \dfrac{1}{ {x}^{5} } = 2 \times 2$

$\bf \implies( x ^{5} + \dfrac{1}{ {x}^{5} } ) = 4 - (x + \dfrac{1}{x} )$

Value of $\bf x + \dfrac{1}{x}$ is given 2

$\bf \implies( x ^{5} + \dfrac{1}{ {x}^{5} } ) = 4 - 2$

$\bf \implies x ^{5} + \dfrac{1}{ {x}^{5} } = 2$

Squaring Both Sides

$\bf ( x ^{5} + \dfrac{1}{ {x}^{5} } ) ^{2} = {2}^{2}$

Using Identity (a+b)² = a² + 2ab + b²

$\bf \longrightarrow ( {x}^{5} ) ^{2} + 2 \times {x}^{5} \times \dfrac{1}{{x}^{5} } + (\dfrac{1}{ {x}^{5}}) ^{2} = 4$

$\bf {x}^{10} + 2 + \dfrac{1}{ {x}^{10} } = 4$

$\bf \implies {x}^{10} + \dfrac{ 1 }{ {x}^{10} } = 4 - 2$

$\bf \implies {x}^{10} + \dfrac{ 1 }{ {x}^{10} } = 2$

$\bf \implies {x}^{10} + \dfrac{ 1 }{ {x}^{10} } = 7-5$

$\bf \implies {x}^{10} + \dfrac{ 1 }{ {x}^{10} } + 5 = 7$