Question

If \[{x^2} + {y^2} + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} = 4\], then the value of \[{x^2} + {y^2}\] is
(a) 2
(b) 4
(c) 8
(d) 16

Answer 1

Hint: Here, we need to find the value of the expression \[{x^2} + {y^2}\]. We will rewrite the given equation and factorise the terms of the expression using the algebraic identity for the square of the difference of two numbers. Then, using the property of the square of a number, we will simplify the expression \[{x^2} + {y^2}\] and find the required value.
Formula Used:
We will use the formula of the square of the difference of two numbers is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\].

Complete step-by-step answer:
First, we will rewrite the given equation.
Subtracting 4 from both sides of the given equation \[{x^2} + {y^2} + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} = 4\], we get
\[ \Rightarrow {x^2} + {y^2} + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} - 4 = 0\]
Rewriting 4 as the sum of 2 and 2, we get
\[\begin{array}{l} \Rightarrow {x^2} + {y^2} + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} - \left( {2 + 2} \right) = 0\\ \Rightarrow {x^2} + {y^2} + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} - 2 - 2 = 0\end{array}\]
Grouping the terms using parentheses, we get
\[ \Rightarrow \left( {{x^2} + \dfrac{1}{{{x^2}}} - 2} \right) + \left( {{y^2} + \dfrac{1}{{{y^2}}} - 2} \right) = 0\]
Now, we will use algebraic identity for square of difference of two numbers to simplify the above expression.
Substituting \[a = x\] and \[b = \dfrac{1}{x}\] in the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], we get
\[ \Rightarrow {\left( {x - \dfrac{1}{x}} \right)^2} = {x^2} + {\left( {\dfrac{1}{x}} \right)^2} - 2\left( x \right)\left( {\dfrac{1}{x}} \right)\]
Simplifying the expression, we get
\[ \Rightarrow {\left( {x - \dfrac{1}{x}} \right)^2} = {x^2} + \dfrac{1}{{{x^2}}} - 2\]
Substituting \[a = y\] and \[b = \dfrac{1}{y}\] in the algebraic identity, we get
\[ \Rightarrow {\left( {y - \dfrac{1}{y}} \right)^2} = {y^2} + {\left( {\dfrac{1}{y}} \right)^2} - 2\left( y \right)\left( {\dfrac{1}{y}} \right)\]
Simplifying the expression, we get
\[ \Rightarrow {\left( {y - \dfrac{1}{y}} \right)^2} = {y^2} + \dfrac{1}{{{y^2}}} - 2\]
Substituting \[{x^2} + \dfrac{1}{{{x^2}}} - 2 = {\left( {x - \dfrac{1}{x}} \right)^2}\] and \[{y^2} + \dfrac{1}{{{y^2}}} - 2 = {\left( {y - \dfrac{1}{y}} \right)^2}\] in the equation \[\left( {{x^2} + \dfrac{1}{{{x^2}}} - 2} \right) + \left( {{y^2} + \dfrac{1}{{{y^2}}} - 2} \right) = 0\], we get
\[ \Rightarrow {\left( {x - \dfrac{1}{x}} \right)^2} + {\left( {y - \dfrac{1}{y}} \right)^2} = 0\]
Now, we know that the square of a number is always greater than or equal to 0.
Therefore, we get
\[ \Rightarrow {\left( {x - \dfrac{1}{x}} \right)^2} \ge 0\] and \[{\left( {y - \dfrac{1}{y}} \right)^2} \ge 0\]
The equation \[{\left( {x - \dfrac{1}{x}} \right)^2} + {\left( {y - \dfrac{1}{y}} \right)^2} = 0\] is true only if \[{\left( {x - \dfrac{1}{x}} \right)^2} = 0\] and \[{\left( {y - \dfrac{1}{y}} \right)^2} = 0\].
Thus, we get
\[ \Rightarrow {\left( {x - \dfrac{1}{x}} \right)^2} = 0\] and \[{\left( {y - \dfrac{1}{y}} \right)^2} = 0\]
Taking the square root of both sides of the equations, we get
\[ \Rightarrow x - \dfrac{1}{x} = 0\] and \[y - \dfrac{1}{y} = 0\]
Rewriting the equations, we get
\[ \Rightarrow x = \dfrac{1}{x}\] and \[y = \dfrac{1}{y}\]
Simplifying the equations by cross-multiplication, we get
\[ \Rightarrow {x^2} = 1\] and \[{y^2} = 1\]
Finally, we can find the value of the expression \[{x^2} + {y^2}\].
Substituting \[{x^2} = 1\] and \[{y^2} = 1\] in the expression \[{x^2} + {y^2}\], we get
\[ \Rightarrow {x^2} + {y^2} = 1 + 1\]
Adding the terms, we get
\[ \Rightarrow {x^2} + {y^2} = 2\]
\[\therefore \] The value of the expression \[{x^2} + {y^2}\] is 2.
Thus, the correct option is option (a).

Note: We factorise the expressions \[\left( {{x^2} + \dfrac{1}{{{x^2}}} - 2} \right)\] and \[\left( {{y^2} + \dfrac{1}{{{y^2}}} - 2} \right)\] using the algebraic identity for square of difference of two numbers. Factorisation is the process of writing an equation as a product of its factors. We factored the algebraic expression \[\left( {{x^2} + \dfrac{1}{{{x^2}}} - 2} \right)\] as \[{\left( {x - \dfrac{1}{x}} \right)^2} = 0\]. This means that \[\left( {x - \dfrac{1}{x}} \right)\] and \[\left( {x - \dfrac{1}{x}} \right)\] are the factors of the algebraic expression \[\left( {{x^2} + \dfrac{1}{{{x^2}}} - 2} \right)\].