Question

If we have an algebraic equation as \[({{x}^{4}}+{{x}^{-4}})=322\], then find, what is one of the values of \[(x-{{x}^{-1}})\]?
A. \[18\]
B. \[16\]
C. \[8\]
D. \[4\]

Answer 1

Hint: In this type of question we have to use algebraic formulae to relate terms.
To find the value of \[(x-{{x}^{-1}})\], first we have to find the value of \[\mathop{x}^{2}+\dfrac{1}{\mathop{x}^{2}}\]. from \[\mathop{x}^{4}+\mathop{x}^{-4}=322\] or \[\mathop{x}^{4}+\dfrac{1}{\mathop{x}^{4}}=322\],
To get \[\mathop{x}^{2}+\dfrac{1}{\mathop{x}^{2}}\], we have to use the algebraic formulae of ${{\left( a+b \right)}^{2}}$ as given below
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
We can rearrange it as
\[\Rightarrow \mathop{a}^{2}+\mathop{b}^{2}=\mathop{(a+b)}^{2}-2ab\]
We can also use algebraic formulae of ${{\left( a-b \right)}^{2}}$ as given below
\[\Rightarrow \mathop{(a-b)}^{2}={{a}^{2}}+{{b}^{2}}-2ab\]
We can rearrange it as
\[\Rightarrow \mathop{a}^{2}+\mathop{b}^{2}=\mathop{(a-b)}^{2}+2ab\].

Complete step-by-step solution -
In the given question we have \[\mathop{x}^{4}+\mathop{x}^{-4}=322\].
We can write it as sum of square of two terms as given below
\[\Rightarrow {{x}^{4}}+{{x}^{-4}}=322\]
$\Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}=322$ $\left\{ \therefore {{x}^{-n}}=\dfrac{1}{{{x}^{n}}} \right\}$
$\Rightarrow {{\left( {{x}^{2}} \right)}^{2}}+\dfrac{1}{{{({{x}^{2}})}^{2}}}=322$
Now we can apply \[{{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab\]
In this \[a={{x}^{2}},b=\dfrac{1}{{{x}^{2}}}\]
\[\Rightarrow \mathop{\left( \mathop{x}^{2}+\dfrac{1}{\mathop{x}^{2}} \right)}^{2}-2\mathop{x}^{2}.\dfrac{1}{\mathop{x}^{2}}=322\]
\[\Rightarrow \mathop{\left( \mathop{x}^{2}+\dfrac{1}{\mathop{x}^{2}} \right)}^{2}-2=322\]
\[\Rightarrow \mathop{\left( \mathop{x}^{2}+\dfrac{1}{\mathop{x}^{2}} \right)}^{2}=322+2\]
\[\Rightarrow \mathop{\left( \mathop{x}^{2}+\dfrac{1}{\mathop{x}^{2}} \right)}^{2}=\sqrt{324}\]
\[\Rightarrow \mathop{x}^{2}+\dfrac{1}{\mathop{x}^{2}}=18\]
\[\therefore \mathop{x}^{2}+\dfrac{1}{\mathop{x}^{2}}=18\]
Now, applying \[{{a}^{2}}+{{b}^{2}}={{(a-b)}^{2}}+2ab\] where $a=x,b=\dfrac{1}{x}$
\[\Rightarrow \mathop{\left( x-\dfrac{1}{x} \right)}^{2}+2x.\dfrac{1}{x}=18\]
\[\Rightarrow \mathop{\left( x-\dfrac{1}{x} \right)}^{2}+2=18\]
\[\Rightarrow \mathop{\left( x-\dfrac{1}{x} \right)}^{2}=18-2\]
\[\Rightarrow x-\dfrac{1}{x}=\sqrt{16}\]
\[\Rightarrow x-\dfrac{1}{x}=4\]
Thus, \[x-\mathop{x}^{-1}=4\]
Therefore the value of \[x-\mathop{x}^{-1}\] is \[4\]
Option 1) \[4\] is the right answer.

Note: On this above method we have to know that formulae properly and first we need to use first \[{{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab\] and then \[{{a}^{2}}+{{b}^{2}}={{(a-b)}^{2}}+2ab\]
If we use first \[{{a}^{2}}+{{b}^{2}}={{(a-b)}^{2}}+2ab\] then we will get value of ${{x}^{2}}-\dfrac{1}{{{x}^{2}}}$ which we can’t use in next formula because for next formula we need value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$
If we didn’t choose the correct order for formula then in the end we will not get the required value.