Question

If \[x + \dfrac{1}{x} = 4\] then \[{x^4} + \dfrac{1}{{{x^4}}}\]\[ = ?\]

Answer 1

Hint: On squaring twice the given expression and simplifying we get the required value. Squaring first and simplifying we get the \[{x^2} + \dfrac{1}{{{x^2}}}\] then on squaring the term again and simplifying we get the value of\[{x^4} + \dfrac{1}{{{x^4}}}\]. The formula for getting square of two term is given as \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] we have to use this formula for squaring in order to get the desired result.

Complete answer:
We have given that and we have to find the value of
We will square the given expression twice and simplify in order to get the result.
On squaring both side of the equation \[x + \dfrac{1}{x} = 4\]
We get,
\[{\left( {x + \dfrac{1}{x}} \right)^2} = {\left( 4 \right)^2}\]
On using the formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
We get,
\[
  {\left( {x + \dfrac{1}{x}} \right)^2} = {\left( x \right)^2} + {\left( {\dfrac{1}{x}} \right)^2} + 2\left( x \right)\left( {\dfrac{1}{x}} \right) \\
   \\
 \]
Similar term in numerator and denominator will get cancelled and so we have,
\[{\left( {x + \dfrac{1}{x}} \right)^2} = {x^2} + \dfrac{1}{{{x^2}}} + 2\]
Substituting it in given equation we have,
\[
  {\left( {x + \dfrac{1}{x}} \right)^2} = {\left( 4 \right)^2} \\
   \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 16 \\
 \]
On subtracting \[2\] both sides we get,
\[ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} = 14\]
For getting the final result we need to square the above expression again.
So, on squaring both side of the equation \[{x^2} + \dfrac{1}{{{x^2}}} = 14\]
We get,
\[{\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = {\left( {14} \right)^2}\]
On using the formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
We get,
\[
  {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = {\left( {{x^2}} \right)^2} + {\left( {\dfrac{1}{{{x^2}}}} \right)^2} + 2\left( {{x^2}} \right)\left( {\dfrac{1}{{{x^2}}}} \right) \\
   \\
 \]
The power of power will get multiplied and similar term in numerator and denominator will get cancelled and so we have,
\[{\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = {x^4} + \dfrac{1}{{{x^4}}} + 2\]
Substituting it in given equation we have,
\[
  {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = {\left( {14} \right)^2} \\
   \Rightarrow {x^4} + \dfrac{1}{{{x^4}}} + 2 = 196 \\
 \]
On subtracting \[2\] both sides we get,
\[ \Rightarrow {x^4} + \dfrac{1}{{{x^4}}} = 194\]

Hence if \[x + \dfrac{1}{x} = 4\] then the value of \[{x^4} + \dfrac{1}{{{x^4}}}\] is \[194\]
So, option (A) is the correct answer.

Note:
\[{\left( {a + b} \right)^2} = \left( {a + b} \right)\left( {a + b} \right)\]. The square of any integer is always positive. However the square root of a positive integer has two values, one is positive and the other one is negative. The power of any number is unique, that is there is only one value of \[{x^a}\] for real \[x\] and integer \[a\]. \[\dfrac{1}{{{x^a}}}\] May also be written as \[{x^{ - a}}\]. The value of a given term is unique.