Answer:

ANSWER:

Given:

(x - 2)/(x + 2) + (x + 2)/(x - 2) = 4

To Find:

Value of x

Solution:

We are given that,

:\implies\dfrac{x-2}{x+2}+\dfrac{x+2}{x-2}=4:⟹

x+2

x−2

+

x−2

x+2

=4

Now we'll take LCM,

:\implies\dfrac{(x-2)(x-2)+(x+2)(x+2)}{(x+2)(x-2)}=4:⟹

(x+2)(x−2)

(x−2)(x−2)+(x+2)(x+2)

=4

So,

:\implies\dfrac{(x-2)^2+(x+2)^2}{(x+2)(x-2)}=4:⟹

(x+2)(x−2)

(x−2)

2

+(x+2)

2

=4

We know that,

:\hookrightarrow(a\pm b)^2=a^2\pm2ab+b^2:↪(a±b)

2

=a

2

±2ab+b

2

And,

:\hookrightarrow(a+b)(a-b)=a^2-b^2:↪(a+b)(a−b)=a

2

−b

2

So,

:\implies\dfrac{(x-2)^2+(x+2)^2}{(x+2)(x-2)}=4:⟹

(x+2)(x−2)

(x−2)

2

+(x+2)

2

=4

:\implies\dfrac{(x^2-4x+4)+(x^2+4x+4)}{(x^2-4)}=4:⟹

(x

2

−4)

(x

2

−4x+4)+(x

2

+4x+4)

=4

Opening brackets,

:\implies\dfrac{x^2-4x+4+x^2+4x+4}{x^2-4}=4:⟹

x

2

−4

x

2

−4x+4+x

2

+4x+4

=4

:\implies\dfrac{2x^2+8}{x^2-4}=4:⟹

x

2

−4

2x

2

+8

=4

Transposing x² - 4 to RHS,

:\implies2\!\!\!/\:(x^2+4)=4\!\!\!/^2\:(x^2-4):⟹2/(x

2

+4)=4/

2

(x

2

−4)

So,

:\implies x^2+4=2(x^2-4):⟹x

2

+4=2(x

2

−4)

Opening the brackets,

:\implies x^2+4=2x^2-8:⟹x

2

+4=2x

2

−8

Transposing LHS to RHS,

:\implies 0=2x^2-8-(x^2+4):⟹0=2x

2

−8−(x

2

+4)

So,

:\implies 0=2x^2-8-x^2-4:⟹0=2x

2

−8−x

2

−4

:\implies x^2-12=0:⟹x

2

−12=0

Transposing 12 to RHS,

:\implies x^2=12:⟹x

2

=12

:\implies x=\pm\sqrt{12}:⟹x=±

12

Therefore,

:\implies\bf x=\pm2\sqrt2:⟹x=±2

2

Hence, the value of x is ±2√2.