Question

Q.2) If (x + 1/x) = 3, then find the value of: a) (x² + 1/x²) and b) (x⁴ + 1/x⁴). (4 marks)​

Answer 1

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{} x + \dfrac{1}{x} = 3 \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{ {x}^{2} + \dfrac{1}{ {x}^{2} } } \\  &\sf{ {x}^{4} + \dfrac{1}{ {x}^{4} } } \end{cases}\end{gathered}\end{gathered}$

$\large\sf{\underbrace{\underline{Formula\ to\ be\ used:-}}}$

$: \implies \tt \: \boxed{ \red{ \bf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\large \underbrace{\underline\purple{\bold{Solution :- }}}$

$\sf{\underbrace{\underline{ \purple{According\ to\ the\ question:-}}}}$

$: \implies \tt \: x \: + \: \dfrac{1}{x} = 3$

★ On squaring both sides, we get

$: \implies \tt \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 9$

$: \implies \tt \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 9$

$: \implies \tt \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 9 - 2$

$: \implies \boxed{ \red{\bf \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 7}}$

★ On squaring both sides, we get

$: \implies \tt \: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 \times {x}^{2} \times \dfrac{1}{ {x}^{2} } = 49$

$: \implies \tt \: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 = 49$

$: \implies \tt \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 49 - 2$

$: \implies \boxed{ \red{\bf \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 47}}$

Answer 2

Answer:

$\displaystyle{\sf\:a\:)\:{\boxed{\red{\sf\:x^2\:+\:\dfrac{1}{x^2}\:=\:7}}}}$

$\displaystyle{\sf\:b\:)\:{\boxed{\red{\sf\:x^4\:+\:\dfrac{1}{x^4}\:=\:47}}}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:\left(\:x\:+\:\dfrac{1}{x}\:\right)\:=\:3}$

We have to find the value of,

a) $\displaystyle{\sf\:\left(\:x^2\:+\:\dfrac{1}{x^2}\:\right)}$

b) $\displaystyle{\sf\:\left(\:x^4\:+\:\dfrac{1}{x^4}\:\right)}$

Now,

$\displaystyle{\sf\:\left(\:x\:+\:\dfrac{1}{x}\:\right)\:=\:3}$

$\displaystyle{\implies\sf\:\left(\:x\:+\:\dfrac{1}{x}\:\right)^2\:=\:(\:3\:)^2\:\quad\:\:\:-\:-\:-\:[\:Squaring\:both\:sides\:]}$

$\displaystyle{\implies\sf\:x^2\:+\:2\:\times\:\cancel{x}\:\times\:\dfrac{1}{\cancel{x}}\:+\:\left(\:\dfrac{1}{x}\:\right)^2\:=\:9}$

$\displaystyle{\implies\sf\:x^2\:+\:2\:+\:\dfrac{1^2}{x^2}\:=\:9\:\quad\:\:-\:-\:-\:\left[\:\because\:\left(\:\dfrac{a}{b}\:\right)^m\:=\:\dfrac{a^m}{b^m}\:\right]}$

$\displaystyle{\implies\sf\:x^2\:+\:2\:+\:\dfrac{1}{x^2}\:=\:9}$

$\displaystyle{\implies\sf\:x^2\:+\:\dfrac{1}{x^2}\:=\:9\:-\:2}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:x^2\:+\:\dfrac{1}{x^2}\:=\:7}}}}$

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Now,

$\displaystyle{\implies\sf\:x^2\:+\:\dfrac{1}{x^2}\:=\:7}$

$\displaystyle{\implies\sf\:\left(\:x^2\:+\:\dfrac{1}{x^2}\:\right)^2\:=\:(\:7\:)^2\:\quad\:\:\:-\:-\:-\:[\:Squaring\:both\:sides\:]}$

$\displaystyle{\implies\sf\:(\:x^2\:)^2\:+\:2\:\times\:\cancel{x^2}\:\times\:\dfrac{1}{\cancel{x^2}}\:+\:\left(\:\dfrac{1}{x^2}\:\right)^2\:=\:49}$

$\displaystyle{\implies\sf\:x^4\:+\:2\:+\:\dfrac{1^4}{x^4}\:=\:9\:\quad\:\:-\:-\:-\:\left[\:\because\:\left(\:\dfrac{a}{b}\:\right)^m\:=\:\dfrac{a^m}{b^m}\:\right]}$

$\displaystyle{\implies\sf\:x^4\:+\:2\:+\:\dfrac{1}{x^4}\:=\:49}$ $\displaystyle{\implies\sf\:x^4\:+\:\dfrac{1}{x^4}\:=\:49\:-\:2}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:x^4\:+\:\dfrac{1}{x^4}\:=\:47}}}}$