Question

The solution set of the equation \[\dfrac{{x - a}}{{x - b}} + \dfrac{{x - b}}{{x - a}} = \dfrac{a}{b} + \dfrac{b}{a}\] , \[a \ne b\] is

Answer 1

Hint: Here, we are given a quadratic equation, with a condition \[a \ne b\]. In this, we will take LCM and use the factorization method to solve this equation. We will also use the identity \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] and then, after that we will use the transposition method and evaluate the given equation and find the values of x. The solution for this equation is the values of x, which are also called zeroes (zeros of polynomials). Thus, they are the solutions for which the equation is satisfied.

Complete step by step solution:
Given that, \[a \ne b\] and \[\dfrac{{x - a}}{{x - b}} + \dfrac{{x - b}}{{x - a}} = \dfrac{a}{b} + \dfrac{b}{a}\]
Taking LCM (x-b) and (x-a) on LHS and ab on RHS.
By using the factorization method, we will get,
\[ \Rightarrow \dfrac{{(x - a)(x - a) + (x - b)(x - b)}}{{(x - a)(x - b)}} = \dfrac{{a(a) + b(b)}}{{ab}}\]
Simplify this above quadratic equation, we will get,
\[ \Rightarrow \dfrac{{{{(x - a)}^2} + {{(x - b)}^2}}}{{(x - a)(x - b)}} = \dfrac{{{a^2} + {b^2}}}{{ab}}\]
We know that, \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] and applying this, we will get,
\[ \Rightarrow \dfrac{{({x^2} - 2xa + {a^2}) + ({x^2} - 2xb + {b^2})}}{{{x^2} - xb - xa + ab}} = \dfrac{{{a^2} + {b^2}}}{{ab}}\]
Removing the brackets, we will get,
\[ \Rightarrow \dfrac{{{x^2} - 2xa + {a^2} + {x^2} - 2xb + {b^2}}}{{{x^2} - ax - bx + ab}} = \dfrac{{{a^2} + {b^2}}}{{ab}}\]
Taking common terms, we get,
\[ \Rightarrow \dfrac{{2{x^2} - 2x(a + b) + {a^2} + {b^2}}}{{{x^2} - (a + b)x + ab}} = \dfrac{{{a^2} + {b^2}}}{{ab}}\]
By using cross multiplication, we get,
\[ \Rightarrow ab(2{x^2} - 2x(a + b) + {a^2} + {b^2}) = ({a^2} + {b^2})({x^2} - (a + b)x + ab)\]
Removing the brackets, we get,
\[ \Rightarrow 2ab{x^2} - 2abx(a + b) + ab({a^2} + {b^2}) = ({a^2} + {b^2}){x^2} - (a + b)({a^2} + {b^2})x + ab({a^2} + {b^2})\]
On evaluating this, we get,
\[ \Rightarrow 2ab{x^2} - 2abx(a + b) = ({a^2} + {b^2}){x^2} - (a + b)({a^2} + {b^2})x\]
By using the transposition method, move all the LHS terms to RHS, we get,
\[ \Rightarrow 0 = ({a^2} + {b^2}){x^2} - (a + b)({a^2} + {b^2})x - 2ab{x^2} + 2abx(a + b)\]
Rearranging this quadratic equation, we get,
\[ \Rightarrow ({a^2} + {b^2}){x^2} - (a + b)({a^2} + {b^2})x - 2ab{x^2} + 2abx(a + b) = 0\]
Taking \[{x^2}\] terms on one side and x terms on other side, we will get,
\[ \Rightarrow ({a^2} + {b^2}){x^2} - 2ab{x^2} - (a + b)({a^2} + {b^2})x + 2abx(a + b) = 0\]
\[ \Rightarrow ({a^2} + {b^2} - 2ab){x^2} - (a + b)({a^2} + {b^2} - 2ab)x = 0\]
We know that, \[{a^2} + {b^2} - 2ab = {(a - b)^2}\]
Applying this, we will get,
\[ \Rightarrow {(a - b)^2}{x^2} - (a + b){(a - b)^2}x = 0\]
\[ \Rightarrow {(a - b)^2}\{ {x^2} - (a + b)x\} = 0\]
First,
\[{(a - b)^2} = 0\]
\[ \Rightarrow (a - b) = 0\]
\[ \Rightarrow a = b\]
Here, this is not possible as \[a \ne b\] is given.
Thus, \[{(a - b)^2}\] is not possible.
Next,
\[ \Rightarrow {x^2} - (a + b)x = 0\]
\[ \Rightarrow x\{ x - (a + b)\} = 0\]
\[ \Rightarrow x = 0\] and \[x - (a + b) = 0\]
\[ \Rightarrow x = 0\] and \[x = a + b\]

Hence, $x = 0$ and $x = a + b$ are the two roots of the given equation.

Note:
Quadratics or quadratic equations can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. The solutions to the quadratic equation are the values of the unknown variable x, which satisfy the equation. These solutions are called roots or zeros of quadratic equations.