Solve for x: $x + \dfrac{1}{x} = \dfrac{{21}}{2}$

Hint:The value of x in $x + \dfrac{1}{x} = \dfrac{{21}}{2}$ can be found first cross multiplying the terms of the equation and then solving the quadratic equation in the variable x. There are various methods that can be employed to solve a quadratic equation like completing the square method, using quadratic formula and by splitting the middle term.Using the quadratic formula gives us the roots of the equation directly with ease.

Complete step by step answer:
In the given question, we are required to solve the equation $x + \dfrac{1}{x} = \dfrac{{21}}{2}$.
So, taking LCM of the terms, we have,
$ \Rightarrow \dfrac{{{x^2} + 1}}{x} = \dfrac{{21}}{2}$
Cross multiplying the terms, we get,
$ \Rightarrow 2\left( {{x^2} + 1} \right) = 21x$
Opening the brackets, we get,
$ \Rightarrow 2{x^2} + 2 = 21x$
Resembling the general form of a quadratic equation $a{x^2} + bx + c = 0$, we get,
$ \Rightarrow 2{x^2} - 21x + 2 = 0$
Comparing with standard quadratic equation $a{x^2} + bx + c = 0$
Here,$a = 2$, $b = - 21$ and$c = 2$.

Now, Using the quadratic formula, we get the roots of the equation as:
$x = \dfrac{{( - b) \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting the values of a, b, and c in the quadratic formula, we get,
$x = \dfrac{{ - \left( { - 21} \right) \pm \sqrt {{{( - 21)}^2} - 4 \times 2 \times \left( 2 \right)} }}{{2 \times 2}}$
Doing the calculations,
$ \Rightarrow x = \dfrac{{21 \pm \sqrt {441 - 16} }}{{2 \times 2}}$
Simplifying the expression, we get,
$ \Rightarrow x = \dfrac{{21 \pm \sqrt {25 \times 7} }}{4}$
$ \Rightarrow x = \dfrac{{21 \pm 5\sqrt 7 }}{4}$
So, $x = \dfrac{{21 + 5\sqrt 7 }}{4}$ and $x = \dfrac{{21 - 5\sqrt 7 }}{4}$ are the roots of the equation $x + \dfrac{1}{x} = \dfrac{{21}}{2}$.

So, the roots of the equation $x + \dfrac{1}{x} = \dfrac{{21}}{2}$ are: $x = \dfrac{{21 + 5\sqrt 7 }}{4}$ and $x = \dfrac{{21 - 5\sqrt 7 }}{4}$.

Note:Quadratic equations are the polynomial equations with degree of the variable or unknown as 2. Quadratic formula is the easiest and most efficient formula to calculate the roots of an equation. Quadratic equations can also be solved by splitting the middle term and completing the square method. Quadratic equations may also be solved by a hit and trial method if the roots of the equation are easy to find.