Question

Solve the following quadratic equations by factorization:
(x+1/x-1)-(x-1/x+1)=5/6,x≠1,-1

Answer 1

x=5 or x= -1/5

Solution:

(x+1)/(x-1)-(x-1)/(x+1)=5/6

=>[(x+1)²-(x-1)²]/(x+1)(x-1)=5/6

=>[(x+1+x-1)(x+1-(x-1))]/x²-1=5/6

=>[(2x)(2)]/x²-1=5/6

=>4x/x²-1=5/6

=>6×(4x)=5(x²-1)

=>24x =5x²-5

=>5x²-24x-5= 0

=>5x²-25x+x-5=0

=>5x(x-5)+1(x-5)=0

=>(x-5)(5x+1)=0

=> x-5=0 , 5x+1=0

=> x=5 , x=-1/5

Answer 2

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

Step-by-step explanation:

The given quadratic equation:

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

To find, the value of x = ?

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

Let $\dfrac{x+1}{x-1}$ = a

The given quadratic equation becomes:

$a-\dfrac{1}{a}=\dfrac{5}{6}$

⇒ $\dfrac{a^2-1}{a}=\dfrac{5}{6}$

By crossmultiplication,

⇒ 6($a^2$ - 1) = 5a

⇒ 6$a^2$ - 5a - 6 = 0

⇒ 6$a^2$ - 9a + 4a - 6 = 0

⇒ 3a(2a - 3) + 2(2a - 3) = 0

⇒ (2a - 3) (3a + 2) = 0

⇒ a = $\dfrac{3}{2}$ or, $\dfrac{-2}{3}$

∴ $\dfrac{x+1}{x-1}$ = $\dfrac{3}{2}$

⇒ 2x + 2 = 3x - 3

⇒ x = 5

Also, $\dfrac{x+1}{x-1}$ = $\dfrac{-2}{3}$

⇒ 3x + 3 = - 2x + 2

⇒ 3x + 2x = 2 - 3

⇒ 5x = - 1

⇒ x = $\dfrac{-1}{5}$

∴ x = 5 or, $\dfrac{-1}{5}$