Question

Find the roots of the folowing equation by Factorisation.
x+1/x-1 - x-1/x+1 =5/6​

Answer 1

Given :

•The given equation is

$\\ \red \bigstar \: \bf \: \frac{x + 1}{x - 1} - \frac{x - 1}{x + 1} = \frac{5}{6}$

To Find :

•The roots of the given equation =?

Solution :

$\\ \bullet \bf \: \frac{x + 1}{x - 1} - \frac{x - 1}{x + 1} = \frac{5}{6}$

$\\ \red\star\blue{\underline{\sf{Getting \: cross \: multiplication :}}}$

$\\ \implies \sf \: \: \frac{x + 1}{x - 1} - \frac{x - 1}{x + 1} = \frac{5}{6}$

$\\ \implies \sf \: \frac{x + 1}{x - 1} - \frac{ x - 1}{x + 1} = \frac{5}{6} \\ \\ \\ \\ \implies \sf \: \frac{(x + 1)(x + 1) - ((x - 1)(x - 1)}{(x - 1)(x + 1)} = \frac{5}{6} \\ \\ \\ \\ \implies \sf \: \frac{x(x + 1) + 1(x + 1) - ((x(x - 1) - 1(x - 1))}{x(x + 1) - 1(x + 1)} = \frac{5}{6} \\ \\ \\ \\ \implies \sf \: \: \frac{ ({x}^{2} + x + x + 1) - ( {x}^{2} - x - x + 1) }{ {x}^{2} + x - x - 1 } = \frac{5}{6} \\ \\ \\ \\ \implies \sf \: \frac{ ({x}^{2} + 2x + 1) - ( {x}^{2} - 2x + 1) }{{x}^{2} - 1} = \frac{5}{6} \\ \\ \\ \\ \implies \sf \: \: \frac{( {x}^{2} + 2x + 1 - {x}^{2} + 2x - 1) }{ {x}^{2} - 1} = \frac{5}{6} \\ \\ \\ \\ \implies \sf \: \frac{4x}{ {x}^{2} - 1} = \frac{5}{6} \\ \\ \\ \implies \sf \: 6(4x) = 5( {x}^{2} - 1) \\ \\ \\ \implies \sf \: 24x = 5 {x}^{2} - 5 \\ \\ \\ \implies \sf \: 24x - 5 {x}^{2} + 5 = 0 \\ \\ \\ \implies \sf \: - 5 {x}^{2} + 24x + 5 = 0 \\ \\ \\ \implies \sf \: 5 {x}^{2} - 24x - 5 = 0 \\ \\ \\ \implies \sf \: 5 {x}^{2} - 25x + x - 5 =0 \\ \\ \\ \implies \sf \:5x(x - 5) + 1(x - 5) = 0 \\ \\ \\ \implies \sf \: (5x + 1) = 0 \: \: \: and \: \: \: (x - 5) = 0 \\ \\ \\ \implies \sf \: \boxed{ \sf{x = \frac{ - 1}{5} }} \: \: \: and \: \: \: \boxed{ \sf{x = 5}}$

The roots of the given quadratic equation is

$\red \bigstar \boxed { \sf{x = \frac{ - 1}{5} \: \: \: and \: \: \: x = 5}}$

$\\ \bigstar\large\green{\underline{\sf{Additional \: information :}}}$

• For solving any type of quadratic equation we use formula method and determinant method to solve easily.

Formula Method :

$\\ \bullet\boxed{\sf{x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}}}$

Determinant Method :

$\\ \boxed{\sf{\bullet \: \delta D=b^{2}-4ac}}$