Question

find the root of the following​

Answer 1

Topic :-

Equations

Given :-

$\dfrac{1}{x+1}+\dfrac{2}{x+2}=\dfrac{4}{x+4}$

To Find :-

Value of 'x' satisfying given equation.

Solution :-

$\dfrac{1}{x+1}+\dfrac{2}{x+2}=\dfrac{4}{x+4}$

Simplifying it,

$\dfrac{1(x+2)+2(x+1)}{(x+1)(x+2)}=\dfrac{4}{x+4}$

$\dfrac{x+2+2x+2}{x(x+2)+1(x+2)}=\dfrac{4}{x+4}$

$\dfrac{3x+4}{x^2+2x+x+2}=\dfrac{4}{x+4}$

$\dfrac{3x+4}{x^2+3x+2}=\dfrac{4}{x+4}$

Cross Multiply,

( 3x + 4 )( x + 4 ) = 4( x² + 3x + 2 )

3x( x + 4 ) + 4( x + 4 ) = 4x² + 12x + 8

3x² + 12x + 4x + 16 = 4x² + 12x + 8

3x² + 16x + 16 = 4x² + 12x + 8

4x² - 3x² + 12x - 16x + 8 - 16 = 0

x² - 4x - 8 = 0

x² - 4x + 4 - 4 - 8 = 0

( x - 2 )² - 12 = 0

( x - 2 )² = 12

Taking Square root both sides,

x - 2 = ± √(12)

x = 2 ± 2√3

Answer :-

So, possible value of 'x' as roots of the equation are :-

• 2 + 2√3

• 2 - 2√3