Question

Q.7
Find the roots of the following equation by Factorisation.
x+1/x-1 - x-1/x+1 = 5/6 ; x is not equal to 1, -1​

Answer 1

We have:

$\dfrac{x+1}{x-1} -\dfrac{x-1}{x+1} =\dfrac{5}{6}$, x ≠ 1 or - 1

We have to find, the value of x.

Solution:

∴ $\dfrac{x+1}{x-1} -\dfrac{x-1}{x+1} =\dfrac{5}{6}$

⇒ $\dfrac{(x+1)^2-(x-1)^2}{(x-1)(x+1)} =\dfrac{5}{6}$

Using the algebraic identity:

$(a+b)^{2} -(a-b)^{2}=4ab$ and

$(a+b)(a-b)=a^{2} -b^{2}$

⇒ $\dfrac{4x}{x^2-1} =\dfrac{5}{6}$

By cross multiplication, we get

5$x^{2}$ - 5 = 24x

⇒ 5$x^{2}$  - 24x - 5 = 0

To solve by factorisation method, we get

5$x^{2}$  - 25x + x - 5 = 0

⇒ 5x(x - 5) + 1(x - 5) = 0

⇒ (x - 5)(5x + 1) = 0

⇒ x - 5 = 0 or, 5x + 1 = 0

⇒ x - 5 = 0 ⇒ x = 5

or,  5x + 1 = 0 ⇒ x = $-\dfrac{1}{5}$

∴ x = 5 or, $-\dfrac{1}{5}$

Answer 2

SOLUTION

TO SOLVE

$\displaystyle \sf{ \frac{x + 1}{x - 1} - \frac{x - 1}{x + 1} = \frac{5}{6} \: \: \: \: \: \: x \ne 1, - 1}$

EVALUATION

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

$\displaystyle \sf{ \implies \: \frac{ { (x + 1)}^{2} - {(x - 1)}^{2} }{(x + 1)(x - 1)} = \frac{5}{6} \: }$

$\displaystyle \sf{ \implies \: \frac{ 4.x.1 }{ {x}^{2} - 1} = \frac{5}{6} \: }$

$\displaystyle \sf{ \implies \: \frac{ 4x }{ {x}^{2} - 1} = \frac{5}{6} \: }$

$\displaystyle \sf{ \implies \: 5 {x}^{2} - 5 = 24x \: }$

$\displaystyle \sf{ \implies \: 5 {x}^{2} - 24x - 5 = 0 \: }$

$\displaystyle \sf{ \implies \: 5 {x}^{2} - 25x + x - 5 = 0 \: }$

$\displaystyle \sf{ \implies \: 5x(x - 5) + 1(x - 5) = 0 \: }$

$\displaystyle \sf{ \implies \: (x - 5) (5x + 1) = 0 \: }$

We know that if product of two real numbers are zero then they are separately zero

$\displaystyle \sf{ \implies \: (x - 5) = 0 \: \: or \: \: (5x + 1) = 0 \: }$

Now

$\displaystyle \sf{ (x - 5) = 0 \: \: gives \: \: x = 5\: }$

$\displaystyle \sf{ (5x + 1) = 0 \: \: gives \: \: x = - \frac{1}{5} }$

Hence the required solution is

$\displaystyle \sf{ x = \: \: 5 \: , - \frac{1}{5} }$

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