Question

(1+1/x+1) (1-1/x-1) =7/8
solve the quadratic equation using factorization
please answer correctly and as soon as possible

Answer 1

Hi...☺

Here is your answer...✌

$(1 + \frac{1}{x + 1} ) \times (1 - \frac{1}{x - 1} ) = \frac{7}{8} \\ \\ \frac{x + 1 + 1}{x + 1} \times \frac{x - 1 - 1}{x - 1} = \frac{7}{8} \\ \\ \frac{x + 2}{x + 1} \times \frac{x - 2}{x - 1} = \frac{7}{8} \\ \\ \frac{{x}^{2} - 2 {}^{2} }{ {x}^{2} - {1}^{2} } = \frac{7}{8} \\ \\ \frac{ {x}^{2} - 4}{ {x}^{2} - 1} = \frac{7}{8} \\ \\ = > 8( {x}^{2} - 4) = 7( {x}^{2} - 1) \\ \\ = > 8 {x}^{2} - 32 = {7}x^{2} - 7 \\ \\ = > 8 {x}^{2} - 7 {x}^{2} - 32 + 7 = 0 \\ \\ = > {x}^{2} - 25 = 0 \\ \\ = > {x}^{2} - {5}^{2} = 0 \\ \\ = > (x + 5)(x - 5) = 0 \\ \\ = > x = - 5 \: \: or \: \: x = 5$

Answer 2

Answer:

In factorization, we write an expression as the product of its factors.

Here, the given expression is,

$(1+\frac{1}{x+1} )(1-\frac{1}{x-1} ) = \frac{7}{8}$

$(\frac{x+1+1}{x+1} )(\frac{x-1-1}{x-1} ) = \frac{7}{8}$

$(\frac{x+2}{x+1} )(\frac{x-2}{x-1} ) = \frac{7}{8}$   ( Since, $(a+b)(a-b)=a^2-b^2$ )

$\frac{x^2-4}{x^2-1}= \frac{7}{8}$

$8(x^2-4)= 7(x^2-1)$  ( By cross multiplication )

$8x^2-32= 7x^2-7$

$x^2-25=0$

$x^2-(5)^2=0$

$(x+5)(x-5)=0$  

Which is the required factorization form,

Now, (x+5)(x-5)=0 ⇒ x+5 = 0 or x -5 =0  ( By zero product property )

⇒ x = -5 or x = 5

Which is the required solution.