Question

If $x + \dfrac{1}{x} = 2$ , then ${x^2} + \dfrac{1}{{{x^2}}} = ?$
A.$2$
B.$0$
C.$4$
D.$6$

Answer 1

In this question we have been given that $x + \dfrac{1}{x} = 2$ . So first of all we will square both the sides of the given equation and then we will break down the left hand side by using the algebraic identity. The formula that we will use here is called the square of the sum of two binomials i.e.
${(a + b)^2} = {a^2} + {b^2} + 2ab$ .

Complete answer:
Here we have $x + \dfrac{1}{x} = 2$ .
On squaring both the sides of the equation, we have :
${\left( {x + \dfrac{1}{x}} \right)^2} = {(2)^2}$ .
Now we will apply the formula here
${(a + b)^2} = {a^2} + {b^2} + 2ab$ .
On comparing, we have
$a = x,b = \dfrac{1}{x}$
By applying the formula we can write
${x^2} + {\left( {\dfrac{1}{x}} \right)^2} + 2(x)\left( {\dfrac{1}{x}} \right) = 4$
On simplifying, it gives:
${x^2} + \dfrac{1}{{{x^2}}} = 4 - 2$
${x^2} + \dfrac{1}{{{x^2}}} = 2$ .
Hence the correct option is (A) $2$ .

Note:
We should note that the formula that we have used above is also called the binomial square.
WE should also know how to derive that formula i.e. ${(a + b)^2}$ can be written as
$(a + b)(a + b)$
On simplifying we can write
$a(a + b) + b(a + b)$
${a^2} + ab + ba + {b^2}$
By grouping the terms together, it gives us
${a^2} + 2ab + {b^2}$ . We can see that it is the required formula.