Question

If \[m-\dfrac{1}{m}=5\]; find the value of \[\to {{m}^{2}}-\dfrac{1}{{{m}^{2}}}\]

Answer 1

Hint: First of all, we have to find the value of \[{{\left( m-\dfrac{1}{m} \right)}^{2}}\],then go for the value of \[\left( {{m}^{2}}+\dfrac{1}{{{m}^{2}}} \right)\]. Then, \[{{m}^{2}}-\dfrac{1}{{{m}^{2}}}\] is to be split into the two factors and then, we just need to put the value of \[\left( m-\dfrac{1}{m} \right)\] and \[\left( m+\dfrac{1}{m} \right)\] into the required expression, and hence we get the result.

Formula used : The formula used in the question is that for the difference of squares of two variables\[{{a}^{2}}-{{b}^{2}}=a+b\bullet a-b\]
We also use square of \[\left( a+b \right)\];
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2\times a\times b\]

Complete step-by-step solution:
Firstly , we start off by simplifying the given expression. This can be simplified by using the formula\[{{a}^{2}}-{{b}^{2}}=a+b\bullet a-b\].
For this, first of all split\[{{m}^{2}}-\dfrac{1}{{{m}^{2}}}\],
After splitting, we get:
\[\left( m+\dfrac{1}{m} \right)\left( m-\dfrac{1}{m} \right)\]
Now we have to find the value of \[\left( m+\dfrac{1}{m} \right)\],
For that, Squaring the expression
 \[\left( m+\dfrac{1}{m} \right)\],
We get :
\[\Rightarrow {{\left( m+\dfrac{1}{m} \right)}^{2}}={{m}^{2}}+\dfrac{1}{{{m}^{2}}}+2\times m\times \dfrac{1}{m}\]
\[\Rightarrow {{\left( m+\dfrac{1}{m} \right)}^{2}}={{m}^{2}}+\dfrac{1}{{{m}^{2}}}+2\]
In order to find this ,we first have to find the value of \[{{m}^{2}}+\dfrac{1}{{{m}^{2}}}\]
To find the value of \[{{m}^{2}}+\dfrac{1}{{{m}^{2}}}\]; we can use the relation
  \[m-\dfrac{1}{m}=5\]
Let us square the value of \[\left( m-\dfrac{1}{m} \right)\]given in the question.
Doing so, we get:
\[
  m-\dfrac{1}{m}=5 \\
 \Rightarrow {{\left( m-\dfrac{1}{m} \right)}^{2}}={{5}^{2}} \\
 \Rightarrow \left( {{m}^{2}}+\dfrac{1}{{{m}^{2}}}-2\times m\times \dfrac{1}{m} \right)=25 \\
 \Rightarrow \left( {{m}^{2}}+\dfrac{1}{{{m}^{2}}}-2 \right)=25 \\
\Rightarrow \left( {{m}^{2}}+\dfrac{1}{{{m}^{2}}} \right)=27 \\
\]
Now putting the value of \[\left( {{m}^{2}}+\dfrac{1}{{{m}^{2}}} \right)\] in the above relation for \[{{\left( m+\dfrac{1}{m} \right)}^{2}}\];
Therefore, we get:
\[\Rightarrow {{\left( m+\dfrac{1}{m} \right)}^{2}}={{m}^{2}}+\dfrac{1}{{{m}^{2}}}+2\]
\[
  \Rightarrow {{\left( m+\dfrac{1}{m} \right)}^{2}}=27+2 \\
 \Rightarrow {{\left( m+\dfrac{1}{m} \right)}^{2}}=29 \\
 \Rightarrow \left( m+\dfrac{1}{m} \right)=\sqrt{29} \\
\]
So now we have both the values of \[\left( m-\dfrac{1}{m} \right)\] and \[\left( m+\dfrac{1}{m} \right)\].
Here, putting the above values in the relation:
\[\left( {{m}^{2}}-\dfrac{1}{{{m}^{2}}} \right)\]
\[
 \Rightarrow \left( m+\dfrac{1}{m} \right)\left( m-\dfrac{1}{m} \right)=\sqrt{29}\times 5 \\
 \Rightarrow \left( m+\dfrac{1}{m} \right)\left( m-\dfrac{1}{m} \right)=5\sqrt{29} \\
\]
\[\Rightarrow \left( {{m}^{2}}-\dfrac{1}{{{m}^{2}}} \right)=5\sqrt{29}\]

Therefore the correct answer is \[ \left( {{m}^{2}}-\dfrac{1}{{{m}^{2}}} \right)=5\sqrt{29}\]

Note: While putting the value of both the spilt term first try to find the value of \[{{m}^{2}}+\dfrac{1}{{{m}^{2}}}\],then go for the value of \[\left( m+\dfrac{1}{m} \right)\].
You can easily find the desired result after knowing these values
Alternative method:
You can directly find the value of \[{{m}^{2}}+\dfrac{1}{{{m}^{2}}}\]
By this formula;
\[{{m}^{2}}+\dfrac{1}{{{m}^{2}}}={{\left( m-\dfrac{1}{m} \right)}^{2}}+2\times m\times \dfrac{1}{m}\]