Question

Find the value of ${x^4} + \dfrac{1}{{{x^4}}}$ if we have given $x + \dfrac{1}{x} = 11$ ?

Answer 1

Hint: We get the required value by squaring twice the given expression and simplifying it. Squaring the term first and simplifying gives us ${x^2} + \dfrac{1}{{{x^2}}}$ , then squaring the term again and simplifying gives us ${x^4} + \dfrac{1}{{{x^4}}}$ . The formula for calculating the square of two terms is \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] . In order to achieve the desired result, we must use this formula for squaring.

Complete step-by-step solution:
We've given $x + \dfrac{1}{x} = 11$ , and now we have to find the value of ${x^4} + \dfrac{1}{{{x^4}}}$
We will obtain the result by squaring the provided expression twice and simplify.
When both sides of the equation $x + \dfrac{1}{x} = 11$ are squared, the result is
${\left( {x + \dfrac{1}{x}} \right)^2} = {11^2}$..........................(1)
We will use the formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]and expand ${\left( {x + \dfrac{1}{x}} \right)^2}$
${\left( {x + \dfrac{1}{x}} \right)^2} = {\left( x \right)^2} + {\left( {\dfrac{1}{x}} \right)^2} + 2\left( x \right)\left( {\dfrac{1}{x}} \right)$
${\left( {x + \dfrac{1}{x}} \right)^2} = {\left( x \right)^2} + {\left( {\dfrac{1}{x}} \right)^2} + 2$
We will substitute it in equation 1, we get
$ \Rightarrow {\left( x \right)^2} + {\left( {\dfrac{1}{x}} \right)^2} + 2 = 121$
We have subtracted 2 from both sides, then we get
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} = 119$
We will again square the both sides,
$ \Rightarrow {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = {119^2}$.................(2)
We will use the formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]and expand it
$ \Rightarrow {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = {x^4} + \dfrac{1}{{{x^4}}} + 2\left( {{x^2}} \right)\left( {\dfrac{1}{{{x^2}}}} \right)$
The power of power will be multiplied, and related terms in the numerator and denominator will be cancelled, giving us the following result:
$ \Rightarrow {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = {x^4} + \dfrac{1}{{{x^4}}} + 2$
We will substitute it in equation 2
$ \Rightarrow {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = {119^2}$
\[ \Rightarrow {x^4} + \dfrac{1}{{{x^4}}} + 2 = {119^2}\]
\[ \Rightarrow {x^4} + \dfrac{1}{{{x^4}}} + 2 = 14161\]
We have subtracted 2 from both side
\[ \Rightarrow {x^4} + \dfrac{1}{{{x^4}}} = 14159\]
The value of ${x^4} + \dfrac{1}{{{x^4}}}$ is 14159 if $x + \dfrac{1}{x} = 11$ .

Note: \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]. Any integer's square is always positive. The square root of a positive integer, on the other hand, has two values, one positive and the other negative. Any number's power is unique, which means that there is only one value of \[{x^a}\] for real x and integer a.
We can also write \[\dfrac{1}{{{x^a}}}\] as \[{x^{ - a}}\].