Question

f it is given that $ \left( x+\dfrac{1}{x} \right)=4 $ , find the value of $ \left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right) $
A. 120
B. 194
C. 160
D. 140

Answer 1

Hint: We need to calculate the sum of both the terms raised to the power ‘4’. For that we will first square it both sides of the equation and solve it so we will get the sum $ \left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right) $ as both the terms in $ \left( x+\dfrac{1}{x} \right) $ are reciprocal of each other so the in the $ '2ab' $ term in $ {{(a+b)}^{2}} $ they will cancel each other and only $ '2' $ will be left and then we will square it both sides again and solve so that we would get the value of the required sum in the similar way.

Complete step-by-step answer:
It is given that: $ \left( x+\dfrac{1}{{{x}^{{}}}} \right)=4 $
We will now square both the sides of the equation and we will get the following:
 $ \begin{align}
  & \Rightarrow {{\left( x+\dfrac{1}{x} \right)}^{2}}={{4}^{2}} \\
 & \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2.x.\dfrac{1}{x}=16 \\
 & \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2=16 \\
 & \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=16-2 \\
\end{align} $
 $ \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=14 $ ........(1)
By first squaring and solving we got the sum of $ \left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right) $ which is the sum of both the terms raised to power ‘2’.
We need the sum of both the terms raised to the power ‘4’ so we will square both sides again and solve which will provide us a sum of $ \left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right) $
Squaring equation (1) we will get:
 $ \begin{align}
  & \Rightarrow {{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}={{14}^{2}} \\
 & \Rightarrow {{({{x}^{2}})}^{2}}+{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{2}}+2.{{x}^{2}}.\dfrac{1}{{{x}^{2}}}=196 \\
 & \Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}+2=196 \\
 & \Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}=196-2 \\
 & \Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}=194 \\
\end{align} $
Hence, $ {{x}^{4}}+\dfrac{1}{{{x}^{4}}}=194 $
So, the correct answer is “Option B”.

Note: We can also directly raise power ‘4’ on both the sides of the equation but for solving by that method we will still require the value of $ \left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right) $ .
This is shown below:
 $ \Rightarrow \left( x+\dfrac{1}{x} \right)=4 $
Raising power ‘4’ on both the sides we will get:
 $ \begin{align}
  & \Rightarrow {{\left( x+\dfrac{1}{x} \right)}^{4}}={{4}^{4}} \\
 & \\
\end{align} $
Opening LHS via binomial theorem we get:
 $ \begin{align}
  & ^{4}{{C}_{0}}.{{x}^{4-0}}.{{\left( \dfrac{1}{x} \right)}^{0}}{{+}^{4}}{{C}_{1}}.{{x}^{4-1}}.{{\left( \dfrac{1}{x} \right)}^{1}}{{+}^{4}}{{C}_{2}}.{{x}^{4-2}}.{{\left( \dfrac{1}{x} \right)}^{2}}{{+}^{4}}{{C}_{3}}.{{x}^{4-3}}.{{\left( \dfrac{1}{x} \right)}^{3}}{{+}^{4}}{{C}_{4}}.{{x}^{4-4}}.{{\left( \dfrac{1}{x} \right)}^{4}}=256 \\
 & \Rightarrow {{x}^{4}}+4{{x}^{2}}+6+\dfrac{4}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}}=256 \\
 & \Rightarrow \left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)+4\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)+6=256 \\
 & \Rightarrow \left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)+4\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)=256-6 \\
 & \Rightarrow \left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}} \right)+4\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)=250 \\
\end{align} $
Therefore, to solve this question we still need the value of $ \left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right) $ .
Therefore, we still would have to square both sides to get its value and it will result in a longer and more tedious solution increasing the scope of mistakes.
Therefore, we solve the question by squaring it and then solving it and then repeating the process.