Question

Solve the equation for $x$: $\dfrac{1}{{x - 2}} + \dfrac{2}{{x - 1}} = \dfrac{6}{x}$, $x \ne 0,1,2$

Answer 1

Hint: To solve the given question, we will first take the LCM on the left hand side of the equation given in the question. After taking the LCM, we will cross multiply the equation. After simplifying the equation, we will get a quadratic equation in $x$. To find the roots of this quadratic equation, we will use the factorization method. In this method, we express $a{x^2} + bx + c = 0$ as the product of two linear factors, say $\left( {px + q} \right)$ and $\left( {rx + s} \right)$, where $p$, $q$, $r$ and $s$ are real numbers such that $p \ne 0$ and $r \ne 0$. We will get linear equations in the form of $px + q = 0$ or, $rx + s = 0$. Solving these linear equations, we get the possible roots of the given quadratic equation as $x = - \dfrac{p}{q}$ and $x = - \dfrac{s}{r}$.

Complete step-by-step solution:
We have, $\dfrac{1}{{x - 2}} + \dfrac{2}{{x - 1}} = \dfrac{6}{x}$
We will take LCM on the left-hand side of the above equation. Thus, we will get the following equation.
$ \Rightarrow \dfrac{{\left( {x - 1} \right) + 2\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} = \dfrac{6}{x}$
On multiplication of terms in brackets, we get
$ \Rightarrow \dfrac{{x - 1 + 2x - 4}}{{{x^2} - 2x - x + 2}} = \dfrac{6}{x}$
$ \Rightarrow \dfrac{{3x - 5}}{{{x^2} - 3x + 2}} = \dfrac{6}{x}$
On cross-multiplication, we get
$ \Rightarrow x\left( {3x - 5} \right) = 6\left( {{x^2} - 3x + 2} \right)$
$ \Rightarrow 3{x^2} - 5x = 6{x^2} - 18x + 12$
Shift all the terms on one side
$ \Rightarrow 6{x^2} - 3{x^2} - 18x + 5x + 12 = 0$
After addition and subtraction of like terms, we get
$ \Rightarrow 3{x^2} - 13x + 12 = 0$
Let us express $3{x^2} - 13x + 12 = 0$ as the product of two linear factors.
$ \Rightarrow 3{x^2} - 9x - 4x + 12 = 0$
$ \Rightarrow 3x\left( {x - 3} \right) - 4\left( {x - 3} \right) = 0$
Now, we have two factors $\left( {x - 3} \right)$ and $\left( {3x - 4} \right)$
$ \Rightarrow \left( {x - 3} \right)\left( {3x - 4} \right) = 0$
\[ \Rightarrow x - 3 = 0\] or $3x - 4 = 0$
Solving these linear equations, we get roots of $3{x^2} - 13x + 12 = 0$
$ \Rightarrow x = 3$ or $x = \dfrac{4}{3}$
Therefore, $x = 3$ or $x = \dfrac{4}{3}$ are the two roots of the given equation.

Note: A quadratic equation of the form $a{x^2} + bx + c = 0$ where $a \ne 0$ can be solved by many methods. Such as completing the square, factorization method, by using the quadratic formula, etc. Doesn’t matter with which method we solve any quadratic equation because the roots of the quadratic equation will remain the same.