Question

If \[x=\dfrac{1}{x-5}\], then find \[{{x}^{2}}-\dfrac{1}{{{x}^{2}}}\]

Answer 1

Hint:First we will change the value of \[x=\dfrac{1}{x-5}\] to

\[x-\dfrac{1}{x}\] and then we square the value of

\[x-\dfrac{1}{x}\] to find the square of its numerical value equivalent. After that using the value of \[x-

\dfrac{1}{x}\] we will find the value of \[x+\dfrac{1}{x}\] by using the formula given below as:

\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\], \[{{\left( a-b

\right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]
and \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]

In the place of \[a\] and \[b\] we will place the value of \[x\] and \[\dfrac{1}{x}\].

Complete step by step solution:
First we assume that the value of the variable \[x\] and \[\dfrac{1}{x}\] be written as \[a\] and \[b\] and then we use the formula: \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] to find the value of \[x-
\dfrac{1}{x}\] but before that we change the value of \[x=\dfrac{1}{x-5}\] to \[x-\dfrac{1}{x}\] by:
\[\Rightarrow x=\dfrac{1}{x-5}\]
\[\Rightarrow x-5=\dfrac{1}{x}\]
\[\Rightarrow x-\dfrac{1}{x}=5\]

Now that we have got the value of \[x-\dfrac{1}{x}\] as \[5\], we will use this value to convert the term from \x-\dfrac{1}{x}\ to \x+\dfrac{1}{x}\ by using the formula of \[{{\left( a-b
\right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] and \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] as:
\[\Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{2}}={{5}^{2}}\]
\[\Rightarrow {{x}^{2}}+{{\left( \dfrac{1}{x} \right)}^{2}}-2=25\]
\[\Rightarrow {{x}^{2}}+{{\left( \dfrac{1}{x} \right)}^{2}}=27\]

After this we will put the value of \[x\] and \[\dfrac{1}{x}\] into \[a\] and \[b\] in the formula of \[{{\left(
a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] as:
\[\Rightarrow {{\left( x+\dfrac{1}{x} \right)}^{2}}={{x}^{2}}+{{\left( \dfrac{1}{x} \right)}^{2}}+2\]
\[\Rightarrow {{\left( x+\dfrac{1}{x} \right)}^{2}}=27+2\]
\[\Rightarrow \left( x+\dfrac{1}{x} \right)=\sqrt{29}\]
Now that we have the values of \[\left( x+\dfrac{1}{x} \right)=\sqrt{29}\] and \[\left( x-\dfrac{1}{x}
\right)=5\], we multiply both the value in the formula of :
\[\Rightarrow {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
\[\Rightarrow \left( x+\dfrac{1}{x} \right)\left( x-\dfrac{1}{x} \right)={{x}^{2}}-\dfrac{1}{{{x}^{2}}}\]

we get the value of \[{{x}^{2}}-\dfrac{1}{{{x}^{2}}}\] as:
\[\Rightarrow {{x}^{2}}-\dfrac{1}{{{x}^{2}}}=\left( x+\dfrac{1}{x} \right)\left( x-\dfrac{1}{x} \right)\]
\[\Rightarrow {{x}^{2}}-\dfrac{1}{{{x}^{2}}}=5\sqrt{29}\]

Note:Students may go wrong while calculating for the value of \[x-\dfrac{1}{x}\], as the question needs to be first converted into simpler form so as to square the term in forming equation of \[{{\left( x-\dfrac{1}{x}
\right)}^{2}}={{5}^{2}}\] and from here the question can be solved easily.