1 x 1 2 x 2 4 x 4

Asked by admin @ in Math viewed by 346 People

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

Answered by admin @


\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})}.


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