Question

Solve by completing square method:1/x -1/(x-2)=3

Answer 1

Solution :-

$\dfrac{1}{x} - \dfrac{1}{x - 2} = 3$

Taking LCM

$\implies \dfrac{1(x - 2) - 1(x)}{x(x - 2)} = 3$

$\implies \dfrac{x - 2 - x}{ {x}^{2} - 2x} = 3$

$\implies \dfrac{- 2}{ {x}^{2} - 2x} = 3$

⇒ - 2 = 3(x² - 2x)

⇒ - 2 = 3x² - 6x

⇒ 3x² - 6x = - 2

Now, solve the equation by completing the square method

Divide throughout by 3

$\implies \dfrac{3 {x}^{2} }{3} - \dfrac{6x}{3} = - \dfrac{2}{3}$

$\implies {x}^{2} - 2x = - \dfrac{2}{3}$

It can be written as

$\implies {x}^{2} - 2(x)(1) = - \dfrac{2}{3}$

Adding 1² on both sides

$\implies {x}^{2} - 2(x)(1) + {1}^{2} = - \dfrac{2}{3} + {1}^{2}$

$\implies (x - 1)^{2} = - \dfrac{2}{3} + 1$

[ Because (a - b)² = a² - 2ab + b² ]

$\implies (x - 1)^{2} = \dfrac{ - 2 + 3}{3}$

$\implies (x - 1)^{2} = \dfrac{1}{3}$

Taking square root on both sides

$\implies \sqrt{ (x - 1)^{2} }= \pm \sqrt{\dfrac{1}{3} }$

$\implies x - 1= \pm \dfrac{1}{ \sqrt{3} }$

$\implies x - 1= \dfrac{1}{ \sqrt{3} } \qquad or \qquad x - 1 = - \dfrac{1}{ \sqrt{3} }$

$\implies x = 1 + \dfrac{1}{ \sqrt{3} } \qquad or \qquad x= 1- \dfrac{1}{ \sqrt{3} }$

$\implies x = \dfrac{ \sqrt{3} + 1}{ \sqrt{3} } \qquad or \qquad x= \dfrac{ \sqrt{3} - 1}{ \sqrt{3} }$

Hence, (√3 + 1)/√3 and (√3 - 1)/√3 are the roots of the equation.