Question

If \[x + \dfrac{1}{x} = 5\], then find the value of ${x^2} + \dfrac{1}{{{x^2}}}$.

Answer 1

Hint: We will first write the identity which we will use in the solution. After that, we will assume a and b to be as we require in the solution. Then, using that we will thus find the answer.

Complete step-by-step answer:
We will use the formula: ${(a + b)^2} = {a^2} + {b^2} + 2ab$.
To use it in the given question, let us assume that $a = x$ and $b = \dfrac{1}{x}$.
Now, let us put these new a and b in the formula we mentioned above.
So, we have: ${\left( {x + \dfrac{1}{x}} \right)^2} = {x^2} + {\left( {\dfrac{1}{x}} \right)^2} + 2.x.\dfrac{1}{x}$.
On simplifying the RHS, we will get:-
$ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^2} = {x^2} + {\left( {\dfrac{1}{x}} \right)^2} + 2$ (Because $x \times \dfrac{1}{x} = 1$)
Now, simplifying it further, we will get:-
$ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^2} = {x^2} + \dfrac{1}{{{x^2}}} + 2$
Now, we are already given the question that \[x + \dfrac{1}{x} = 5\]. So, putting this in the above equation, we will get:-
$ \Rightarrow {\left( 5 \right)^2} = {x^2} + \dfrac{1}{{{x^2}}} + 2$
Simplifying the LHS, we will get:-
$ \Rightarrow 25 = {x^2} + \dfrac{1}{{{x^2}}} + 2$
Arranging the terms from left to right and right to left, we will get:-
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 25$
Taking the 2 from LHS to RHS, we will get:-
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} = 25 - 2$
Simplifying the values on RHS, we will get:-
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} = 23$.

Hence, the answer is 23.

Note: The students noticed that we did cross $x$ and $\dfrac{1}{x}$ with each other to get 1. This may seem to be simple enough to cross the terms but you must know that you can only cancel terms like this, only when that quantity which is being cut off is never equal to 0. Here, in this question, we knew that it is not possible, because \[x + \dfrac{1}{x} = 5\] is given to us already and this is only possible to be defined when $x \ne 0$ because $\dfrac{1}{x}$ cannot be defined if $x = 0$.
The students must wonder how the identity or formula that we use did come in the picture. We know the binomial theorem which states that ${(a + b)^n}{ = ^n}{C_0}{a^n}{b^0}{ + ^n}{C_1}{a^n}{b^1} + .......{ + ^n}{C_{n - 1}}{a^1}{b^{n - 1}}{ + ^n}{C_n}{a^0}{b^n}$
Now, put n = 2, we will get:-
${(a + b)^2}{ = ^2}{C_0}{a^2}{b^0}{ + ^2}{C_1}{a^1}{b^1}{ + ^2}{C_2}{a^0}{b^2}$ …….(1)
Now, we will use $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$.
So, we will get:-
$^2{C_0} = \dfrac{{2!}}{{0!(2)!}} = 1{,^2}{C_1} = \dfrac{{2!}}{{1!(1)!}} = 2{,^2}{C_2} = \dfrac{{2!}}{{2!(0)!}} = 1$
Putting all these in (1), we will get:-
${(a + b)^2} = {a^2} + 2ab + {b^2}$
Thus, we got our formula.