Question

Solve the following quadratic equation for its roots:
$\dfrac{1}{x} - \dfrac{1}{{x - 2}} = 3$, $x \ne 0,2$

Answer 1

Hint: To solve the given equation, we rearrange the equation such that the denominator terms are eliminated. Then we express the equation in form of ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ to simplify it and solve for x.

Complete step-by-step answer:
Given data,
$ \Rightarrow \dfrac{1}{x} - \dfrac{1}{{x - 2}} = 3$
Taking L.C.M as $\left( x \right)\left( {x - 2} \right)$
$ \Rightarrow \dfrac{{x - 2 - x}}{{x\left( {x - 2} \right)}} = 3$
$ \Rightarrow x - 2 - x = 3x\left( {x - 2} \right)$
$ \Rightarrow - 2 = 3{x^2} - 6x$
$ \Rightarrow 3{x^2} - 6x + 2 = 0$ ……….. (i)
Here, we can see that we cannot make factors of this equation.
So, finding the value of x in this equation by using factorization method is not possible.
So, we will solve this equation by another method.
As we know that, the formula of finding the roots of quadratic equation $a{x^2} + bx + c = 0$ is given by, $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, provided ${b^2} - 4ac \geqslant 0$
Comparing equation (i) with $a{x^2} + bx + c = 0$, we will get
a = 3, b = -6 and c = 2,
putting these values in the above formula, we will get
$ \Rightarrow x = \dfrac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( 3 \right)\left( 2 \right)} }}{{2\left( 3 \right)}}$
$ \Rightarrow x = \dfrac{{6 \pm \sqrt {36 - 24} }}{6}$
$ \Rightarrow x = \dfrac{{6 \pm 2\sqrt 3 }}{6}$
Taking 2 as common,
$ \Rightarrow x = \dfrac{{3 \pm \sqrt 3 }}{3}$
We get two values of x,
i.e. $x = \dfrac{{3 + \sqrt 3 }}{3}$ or $x = \dfrac{{3 - \sqrt 3 }}{3}$
we get,
$x = 1 + \dfrac{1}{{\sqrt 3 }}$,
Or
$x = 1 - \dfrac{1}{{\sqrt 3 }}$
Hence these are the solutions to the given quadratic equation.

Note: In order to solve this type of questions the key is to convert the equation we have in such a way that it can be simplified. We could use a formula or add and subtract an extra term or anything of that sort. And while solving a term which is in the form of a square, we have to remember it can take both positive and negative values, it can give different answers.