Question

Solve the equation by the method of your choice. StartFraction 1 Over x EndFraction plus StartFraction 1 Over x plus 4 EndFraction equals one half The solution set is ​{ nothing​}.

Answer 1

$The\ domain:\\x\neq0\ \vedge\ x+4\neq0\\\\D:x\in\mathbb{R}-\{-4;\ 0\}$

$\dfrac{1}{x}+\dfrac{1}{x+4}=\dfrac{1}{2}\\\\\dfrac{x+4}{x(x+4)}+\dfrac{x}{x(x+4)}=\dfrac{1}{2}\\\\\dfrac{x+4+x}{x^2+4x}=\dfrac{1}{2}\\\\\dfrac{2x+4}{x^2+4x}=\dfrac{1}{2}\ \ \ \ |cross\ multiply\\\\1(x^2+4x)=2(2x+4)\\\\x^2+4x=4x+8\ \ \ \ |-4x\\\\x^2=8\to x=\pm\sqrt8\\\\x=-\sqrt{4\cdot2}\ \vee\ x=\sqrt{4\cdot2}\\\\\boxed{x=-2\sqrt2\ \vee\ x=2\sqrt2}\\\\The\ solution\ set\ is\ \{-2\sqrt2;\ 2\sqrt2\}$

Answer 2

Answer:

$x=2\sqrt{2}$ and $x=2\sqrt{2}$

Step-by-step explanation:

The given equation is

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

We need to solve the given equation.

Taking LCM on left side.

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

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

On cross multiplication we get

$2(2x+4)=x^2+4x$

$4x+8=x^2+4x$

Subtract 4x from both sides.

$8=x^2$

Taking square root on both sides.

$\pm\sqrt{8}=x$

$\pm2\sqrt{2}=x$

Therefore, solutions of the given equation are $x=2\sqrt{2}$ and $x=2\sqrt{2}$.