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# If $x-\dfrac{1}{x}=2$, then the value of ${{x}^{4}}+\dfrac{1}{{{x}^{4}}}$ is?

Last updated date: 21st Jul 2024
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Hint: To solve the above problem, first of all, we are going to square on both the sides of the given equation $x-\dfrac{1}{x}=2$. After solving this equation, we will get the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$ then again we will square on both the sides of the equation and this is how we will get the value of ${{x}^{4}}+\dfrac{1}{{{x}^{4}}}$.

The equation given in the above problem is as follows:
$x-\dfrac{1}{x}=2$
Taking square on both the sides of the above equation we get,
\begin{align} & {{\left( x-\dfrac{1}{x} \right)}^{2}}={{\left( 2 \right)}^{2}} \\ & \Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{2}}=4 \\ \end{align}
We know that the algebraic identity which states that:
${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Substituting $a=x,b=\dfrac{1}{x}$ in the above equation we get,
\begin{align} & {{\left( x-\dfrac{1}{x} \right)}^{2}}={{x}^{2}}-2x\left( \dfrac{1}{x} \right)+{{\left( \dfrac{1}{x} \right)}^{2}} \\ & \Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{2}}={{x}^{2}}-2+\dfrac{1}{{{x}^{2}}} \\ \end{align}
Equating the above equation to 4 we get,
\begin{align} & \Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{2}}={{x}^{2}}-2+\dfrac{1}{{{x}^{2}}}=4 \\ & \Rightarrow {{x}^{2}}-2+\dfrac{1}{{{x}^{2}}}=4 \\ \end{align}
Adding 2 on both the sides of the above equation we get,
\begin{align} & \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=4+2 \\ & \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=6 \\ \end{align}
Now, again, we are going to take square on both the sides of the above equation we get,
\begin{align} & \Rightarrow {{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}={{\left( 6 \right)}^{2}} \\ & \Rightarrow {{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}=36 \\ \end{align}
We are going to use the algebraic identity which states that:
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Substituting $a={{x}^{2}},b=\dfrac{1}{{{x}^{2}}}$ in the above equation we get,
\begin{align} & {{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}={{\left( {{x}^{2}} \right)}^{2}}+2{{x}^{2}}\left( \dfrac{1}{{{x}^{2}}} \right)+{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{2}} \\ & \Rightarrow {{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}={{x}^{4}}+2+\dfrac{1}{{{x}^{4}}} \\ \end{align}
Equating 36 to the above equation we get,
\begin{align} & \Rightarrow {{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}={{x}^{4}}+2+\dfrac{1}{{{x}^{4}}}=36 \\ & \Rightarrow {{x}^{4}}+2+\dfrac{1}{{{x}^{4}}}=36 \\ \end{align}
Subtracting 2 on both the sides of the above equation we get,
\begin{align} & {{x}^{4}}+\dfrac{1}{{{x}^{4}}}=36-2 \\ & \Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}=34 \\ \end{align}
Hence, we have found the value of ${{x}^{4}}+\dfrac{1}{{{x}^{4}}}$ as 34.

Note: To solve the above problem, you must know the algebraic identities of ${{\left( a-b \right)}^{2}}$ and ${{\left( a+b \right)}^{2}}$. If you don’t know this property then you cannot move further in this problem. Also, make sure you have done the calculations correctly and also be mindful while taking the square of ${{x}^{2}}$ because you might incorrectly take the square and hence end up in the wrong form of the expression.