Solve for x:$\frac{1}{x+1} +\frac{2}{x+2} =\frac{4}{x+4} ;\ x\neq -1,\ -2,\ -4$

Given: The equation :$\frac{1}{x+1} +\frac{2}{x+2} =\frac{4}{x+4} ;\ x\
eq -1,\ -2,\ -4$

To do: To solve the given quadratic equation.

Solution:
L.C.M. of all the denominators is $( x\ +\ 1)( x\ +\ 2)( x\ +\ 4)$

Multiply throughout by the L.C.M., we get

$( x\ +\ 2)( x\ +\ 4) \ +\ 2( x\ +\ 1)( x\ +\ 4) \ =\ 4( x\ +\ 1)( x\ +\ 2)$

$( x\ +\ 4)( x\ +\ 2+\ 2x\ +2) \ =\ 4( x^{2} +\ 3x\ +\ 2)$

$( x\ +\ 4)( 3x\ +\ 4) \ =\ 4x^{2} +\ 12x\ +\ 8$

$3x^{2} +\ 16x\ +\ 16\ =\ 4x^{2} +\ 12x\ +\ 8$

$x^{2} -4x-8=0$

here $a=1,\ b=-4$ and $c=-8$

$x=\frac{-b\pm \sqrt{b^{2} -4ac}}{2a} =\frac{-( -4) \pm \sqrt{( -4)^{2}-4\times 1\times ( -8)}}{2\times 1}$

$=\frac{4\pm \sqrt{16+32}}{2}$

$=\frac{4\pm \sqrt{48}}{2}$

$=\frac{4\pm 4\sqrt{3}}{2}$

$=\frac{2( 2\pm 2\sqrt{3})}{2}$

$=2\pm 2\sqrt{3}$
 

Hence, there are two solution of the given equation, $x=2+2\sqrt{3}$ and $x=2-2\sqrt{3}$

Updated on: 10-Oct-2022

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