Question

solve for x 1/x+1 +2/x+2=4/x+4

Answer 1

$\bold{x=(2 \pm \sqrt{3})}$ is the value of x for $\bold{\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4}}$

Given:

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

To find:

Value of x =?

Solution:

To find the value of x, we simplify the following solution:  

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

$\frac{1}{x+1}+\frac{2}{x+2}-\frac{4}{x+4}=0$

Taking the LCM of (x+1)(x+2)(x+4) we find the value of the question in simple form:

$\frac{(x+2)(x+4)+2(x+1)(x+4)-4(x+1)(x+2)}{(x+1)(x+2)(x+4)}=0$

Transferring (x+1)(x+2)(x+4) over to zero to reduce it to zero.

(x+2)(x+4)+2(x+1)(x+4)-4(x+1)(x+2)=0  

Multiplying and subtracting the extra parts

$\begin{array}{l}{x^{2}+6 x+8+2 x^{2}+10 x+8=4 x^{2}+12 x+8} \\ {x^{2}-4 x-8=0}\end{array}$.

As we can see that the equation cannot be factorize using rational number. Hence the values of the roots are:

$\begin{array}{l}{\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}} \\ {x^{2}-4 x-8=0}\end{array}$

The value of a = 1, b = -4 and c = -8

$\begin{array}{l}{=\frac{-(-4) \pm \sqrt{16-4(1)(-8)}}{2}} \\ {(x-(2+\sqrt{3}))(x+(2-\sqrt{3}))}\end{array}$

Therefore, the value of x is  $\bold{x=(2 \pm \sqrt{3})}$.

Answer 2

Answer:

$x=2+2\sqrt{3}\:Or \: x=2-2\sqrt{3}$

Step-by-step explanation:

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

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

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

$\implies \frac{3x+4}{x^{2}+3x+2}=\frac{4}{x+4}$

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

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

$\implies 3x^{2}+16x+16-4x^{2}-12x-8=0$

$\implies -x^{2}+4x+8=0$

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

$Compare \: this \: with \\ax^{2}+bx+c=0,we \:get$

a=1,b=-4,x=-8

Discreminant (D)=b²-4ac

= (-4)²-4×1×(-8)

= 16+32

= 48

$x = \frac{-b±\sqrt{D}}{2a}$

$=\frac{-(-4)±\sqrt{48}}{2}\\=\frac{4±4\sqrt{3}}{2}\\=2±2\sqrt{3}$

$\implies x=2+2\sqrt{3}\:Or \: x=2-2\sqrt{3}$

Therefore,

$x=2+2\sqrt{3}\:Or \: x=2-2\sqrt{3}$

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