Question

If \[{x^2} + \dfrac{1}{{{x^2}}} = 38\], then the value of \[x - \dfrac{1}{x}\].

Answer 1

Hint: Here in this question, we have to find the value of the algebraic expression \[x - \dfrac{1}{x}\] using the given algebraic expression \[{x^2} + \dfrac{1}{{{x^2}}} = 38\], Firstly, we have to add or subtract to the some number and later by using the algebraic identity i.e., \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\] on simplification we get the required value of \[x - \dfrac{1}{x}\].

Complete step by step answer:
A algebraic expression is an expression which is made up of variables and constants, along with algebraic operations (addition, subtraction, etc.). Expressions are made up of terms. They are also termed algebraic equations.
Consider the given Algebraic expression
\[ \Rightarrow \,\,{x^2} + \dfrac{1}{{{x^2}}} = 38\] --------(1)
Now we have to find the value of algebraic expression \[x - \dfrac{1}{x}\] using the above given expression.
Subtracting 2 on both sides in equation (1), then
\[ \Rightarrow \,\,{x^2} + \dfrac{1}{{{x^2}}} - 2 = 38 - 2\]
On simplification we get
\[ \Rightarrow \,\,{x^2} + \dfrac{1}{{{x^2}}} - 2 = 36\] --------(2)
Equation (2) also can be written as
\[ \Rightarrow \,\,{x^2} + \dfrac{1}{{{x^2}}} - 2 \cdot x \cdot \dfrac{1}{x} = 36\]
Because the value of \[2 \cdot x \cdot \dfrac{1}{x} = 2\] hence, by adding this there is no change in the value of expression.
Or
\[ \Rightarrow \,\,{\left( x \right)^2} + {\left( {\dfrac{1}{x}} \right)^2} - 2 \cdot x \cdot \dfrac{1}{x} = 36\]------(3)
Now, we have to compare the LHS of the above expression to the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
Where \[a = x\] and \[b = \dfrac{1}{x}\] then by identity The LHS of equation (3) can be written as
\[ \Rightarrow \,\,{\left( {x - \dfrac{1}{x}} \right)^2} = 36\]
Taking square root on both sides, then
\[ \Rightarrow \,\,\sqrt {{{\left( {x - \dfrac{1}{x}} \right)}^2}} = \pm \sqrt {36} \]
On cancelling square and root we get
\[ \Rightarrow \,\,\left( {x - \dfrac{1}{x}} \right) = \pm \sqrt {36} \]
As we know the 36 is the square number of 6
\[ \Rightarrow \,\,\left( {x - \dfrac{1}{x}} \right) = \pm \sqrt {36} \]
\[ \Rightarrow \,\,\left( {x - \dfrac{1}{x}} \right) = \pm \sqrt {{6^2}} \]
On simplification, we get
\[ \Rightarrow \,\,\left( {x - \dfrac{1}{x}} \right) = \pm \,6\]

Hence, the value of the algebraic expression \[x - \dfrac{1}{x}\] is +6 or -6.

Note: The equation is of the form of algebraic equation. We use simple mathematical operations to simplify. The question contains the square function but we have to find the term which does not contain the square. Therefore we use the square root property to simplify.