Question

Find the roots of the quadratic equation \[\dfrac{1}{x} - \dfrac{1}{{\left( {x - 2} \right)}} = 3\].

Answer 1

Hint: The given problem requires us to solve a quadratic equation using a quadratic formula. We first simplify the equation by taking LCM of both the terms and then form a 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 \[\dfrac{1}{x} - \dfrac{1}{{\left( {x - 2} \right)}} = 3\] .
Firstly, taking the LCM of the two rational terms in variable x, we get,
\[ \Rightarrow \dfrac{{\left( {x - 2} \right) - x}}{{x\left( {x - 2} \right)}} = 3\]
Now, simplifying the quadratic equation by cross multiplying the terms, we get,
\[ \Rightarrow - 2 = 3x\left( {x - 2} \right)\]
Opening the brackets, we get,
\[ \Rightarrow - 2 = 3{x^2} - 6x\]
The quadratic formula can be employed for solving an equation only if we compare the given equation with the standard form of quadratic equation.
Comparing with standard quadratic equation $ a{x^2} + bx + c = 0 $ , we get,
\[ \Rightarrow 3{x^2} - 6x + 2 = 0\]
Here, $ a = 3 $ , $ b = - 6 $ 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,
 $ \Rightarrow x = \dfrac{{ - \left( { - 6} \right) \pm \sqrt {{{( - 6)}^2} - 4 \times 3 \times \left( 2 \right)} }}{{2 \times 3}} $
 $ \Rightarrow x = \dfrac{{6 \pm \sqrt {36 - 24} }}{6} $
Simplifying the calculations, we get,
 $ \Rightarrow x = \dfrac{{6 \pm \sqrt {12} }}{6} $
 $ \Rightarrow x = \dfrac{{6 \pm 2\sqrt 3 }}{6} $
Now, cancelling the common factors in numerator and denominator, we get,
 $ \Rightarrow x = \dfrac{{3 \pm \sqrt 3 }}{3} $
 $ \Rightarrow x = 1 \pm \dfrac{1}{{\sqrt 3 }} $
So, \[x = 1 + \dfrac{1}{{\sqrt 3 }}\] and \[x = 1 - \dfrac{1}{{\sqrt 3 }}\] are the roots of the quadratic equation \[\dfrac{1}{x} - \dfrac{1}{{\left( {x - 2} \right)}} = 3\].
So, the roots of the equation \[\dfrac{1}{x} - \dfrac{1}{{\left( {x - 2} \right)}} = 3\] are: \[x = 1 + \dfrac{1}{{\sqrt 3 }}\] and \[x = 1 - \dfrac{1}{{\sqrt 3 }}\].
So, the correct answer is “\[x = 1 + \dfrac{1}{{\sqrt 3 }}\] and \[x = 1 - \dfrac{1}{{\sqrt 3 }}\]”.

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. We first convert the quadratic equation into the standard form before applying the quadratic formula for finding the roots.