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\\eq0\ \vedge\ x+4\\eq0\\\\D:x\in\mathbb{R}-\{-4;\ 0\}`


`(1)/(x)+(1)/(x+4)=(1)/(2)\\\\(x+4)/(x(x+4))+(x)/(x(x+4))=(1)/(2)\\\\(x+4+x)/(x^2+4x)=(1)/(2)\\\\(2x+4)/(x^2+4x)=(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=-√(4\cdot2)\ \vee\ x=√(4\cdot2)\\\\\boxed{x=-2\sqrt2\ \vee\ x=2\sqrt2}\\\\The\ solution\ set\ is\ \{-2\sqrt2;\ 2\sqrt2\}`

Answer 2

`x=2√(2)` and
`x=2√(2)`

Explanation:

The given equation is

`(1)/(x)+(1)/(x+4)=(1)/(2)`

We need to solve the given equation.

Taking LCM on left side.

`((x+4)+x)/(x(x+4))=(1)/(2)`

`(2x+4)/(x^2+4x)=(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√(8)=x`

`\pm2√(2)=x`

Therefore, solutions of the given equation are
`x=2√(2)` and
`x=2√(2)`.