Question

If an algebraic equation is given as ${{x}^{4}}+\dfrac{1}{{{x}^{4}}}=727$ , then $x-\dfrac{1}{x}$ is equal to.
(a) $5$
(b) $\sqrt{29}$
(c) $25$
(d) $27$

Answer 1

Hint: For solving this question, we will assume $x-\dfrac{1}{x}=t$ , then we will use the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to write the expression of ${{t}^{2}}$ . After that, we will use the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to get an equation between ${{t}^{2}}$ and ${{x}^{4}}+\dfrac{1}{{{x}^{4}}}$ . Then, we will substitute the value of ${{x}^{4}}+\dfrac{1}{{{x}^{4}}}=727$ from the given data and solve for the value of $t$ and select the correct option.

Complete step-by-step solution -
Given:
It is given that, if ${{x}^{4}}+\dfrac{1}{{{x}^{4}}}=727$ and we have to find the value of $x-\dfrac{1}{x}$ .
Now, before we proceed we should know the following formula:
$\begin{align}
  & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab............\left( 1 \right) \\
 & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab............\left( 2 \right) \\
\end{align}$
Now, we will use the above two formulas to solve this question.
Now, let the value of $x-\dfrac{1}{x}=t$ and we can write the expression of ${{\left( x-\dfrac{1}{x} \right)}^{2}}$ using the formula from the equation (1). Then,
$\begin{align}
  & t=x-\dfrac{1}{x} \\
 & \Rightarrow {{t}^{2}}={{\left( x-\dfrac{1}{x} \right)}^{2}} \\
 & \Rightarrow {{t}^{2}}={{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2\times x\times \dfrac{1}{x} \\
 & \Rightarrow {{t}^{2}}={{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2 \\
 & \Rightarrow {{t}^{2}}+2={{x}^{2}}+\dfrac{1}{{{x}^{2}}} \\
\end{align}$
Now, we will write the expression of ${{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}$ using the formula from the equation (2). Then,
$\begin{align}
  & {{t}^{2}}+2={{x}^{2}}+\dfrac{1}{{{x}^{2}}} \\
 & \Rightarrow {{\left( {{t}^{2}}+2 \right)}^{2}}={{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}} \\
 & \Rightarrow {{\left( {{t}^{2}}+2 \right)}^{2}}={{x}^{4}}+\dfrac{1}{{{x}^{4}}}+2\times {{x}^{2}}\times \dfrac{1}{{{x}^{2}}} \\
 & \Rightarrow {{\left( {{t}^{2}}+2 \right)}^{2}}={{x}^{4}}+\dfrac{1}{{{x}^{4}}}+2 \\
\end{align}$
Now, as it is given that value of ${{x}^{4}}+\dfrac{1}{{{x}^{4}}}=727$ so, we can substitute its given value in the above equation. Then,
$\begin{align}
  & {{\left( {{t}^{2}}+2 \right)}^{2}}={{x}^{4}}+\dfrac{1}{{{x}^{4}}}+2 \\
 & \Rightarrow {{\left( {{t}^{2}}+2 \right)}^{2}}=727+2 \\
 & \Rightarrow {{\left( {{t}^{2}}+2 \right)}^{2}}=729 \\
 & \Rightarrow {{\left( {{t}^{2}}+2 \right)}^{2}}={{\left( 27 \right)}^{2}} \\
\end{align}$
Now, we will take the square root of terms on the left-hand side and right-hand side in the above equation. Then,
$\begin{align}
  & {{\left( {{t}^{2}}+2 \right)}^{2}}={{\left( 27 \right)}^{2}} \\
 & \Rightarrow {{t}^{2}}+2=\sqrt{{{\left( 27 \right)}^{2}}} \\
 & \Rightarrow {{t}^{2}}+2=\pm 27 \\
\end{align}$
Now, as we know that for any $t\in R$ value of ${{t}^{2}}+2>0$ so, ${{t}^{2}}+2\ne -27$ . Then,
$\begin{align}
  & {{t}^{2}}+2=27 \\
 & \Rightarrow {{t}^{2}}=27-2 \\
 & \Rightarrow {{t}^{2}}=25 \\
 & \Rightarrow {{t}^{2}}={{5}^{2}} \\
\end{align}$
Now, we will take the square root of terms on the left-hand side and right-hand side in the above equation. Then,
$\begin{align}
  & {{t}^{2}}={{5}^{2}} \\
 & \Rightarrow t=\pm 5 \\
\end{align}$
Now, from the above result, we conclude that value of $t=-5,5$ and as per our assumption $x-\dfrac{1}{x}=t$ . Then,
$\left( x-\dfrac{1}{x} \right)=\pm 5$
Now, from the above result, we conclude that, if ${{x}^{4}}+\dfrac{1}{{{x}^{4}}}=727$ , then the value of $\left( x-\dfrac{1}{x} \right)=\pm 5$ .
Now, we will select the most suitable option from the given options. And it is evident that option (a) is the most suitable option.
Hence, (a) will be the correct option.

Note: Here, the student should first understand what is asked in the problem and then proceed in the right direction to get the correct answer quickly. And in such questions just apply the whole square formula for ${{\left( a+b \right)}^{2}}$ , ${{\left( a-b \right)}^{2}}$ correctly without any mistake sequentially. Moreover, we could have also solved this problem in a reverse manner by writing ${{x}^{4}}+\dfrac{1}{{{x}^{4}}}={{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}-2$ . After that, we will write ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}={{\left( x-\dfrac{1}{x} \right)}^{2}}+2$ and solved further for the value of $\left( x-\dfrac{1}{x} \right)$ easily.