Question

If $x-\dfrac{1}{x}=1$ then what is the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$?
A. $1$
B. $-1$
C. $2$
D. $3$

Answer 1

Hint: To obtain the value of the given algebraic expression we will use the algebraic identity. Firstly we have to find the value of the algebraic expression with a variable with highest power as 2 so we will use ${{\left( a-b \right)}^{2}}$ formula. Then by simplifying the obtain equation we will find the desired answer.

Complete step by step answer:
 It is given to us that the below expression has a value:
$x-\dfrac{1}{x}=1$…..$\left( 1 \right)$
We have to find the value of algebraic expression given below:
${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$…..$\left( 2 \right)$
So as we can see above value has ${{x}^{2}}$ in it so we will use below identity in expression (1):
${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
So find the square of equation (1) by using the above formula as follows:
$\begin{align}
  & {{\left( x-\dfrac{1}{x} \right)}^{2}}={{\left( 1 \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}-2\times x\times \dfrac{1}{x}+{{\left( \dfrac{1}{x} \right)}^{2}}=1 \\
 & \Rightarrow {{x}^{2}}-2+\dfrac{1}{{{x}^{2}}}=1 \\
 & \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=1+2 \\
\end{align}$
On simplifying it we get,
$\therefore {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=3$

So, the correct answer is “Option D”.

Note: An expression made up of variables and constant along with the algebraic operations such as addition, subtraction, multiplication and division is known as algebraic expressions. Algebraic equations on the other hand is an equation in which two expressions are set equal to each other. . The main difference between the two is that algebraic expressions don’t have equal sign and algebraic equations have equal sign. Then we have the algebraic identity that is an algebraic equation which is valid for any value of the variable in it. It is a standard form which can be used for any value of the variables in it. These identities are used to solve many different polynomials and the algebraic equations.