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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​}.

Respuesta :

gmany
[tex]The\ domain:\\x\neq0\ \vedge\ x+4\neq0\\\\D:x\in\mathbb{R}-\{-4;\ 0\}[/tex]

[tex]\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\}[/tex]

Answer:

[tex]x=2\sqrt{2}[/tex] and [tex]x=2\sqrt{2}[/tex]

Step-by-step explanation:

The given equation is

[tex]\frac{1}{x}+\frac{1}{x+4}=\frac{1}{2}[/tex]

We need to solve the given equation.

Taking LCM on left side.

[tex]\frac{(x+4)+x}{x(x+4)}=\frac{1}{2}[/tex]

[tex]\frac{2x+4}{x^2+4x}=\frac{1}{2}[/tex]

On cross multiplication we get

[tex]2(2x+4)=x^2+4x[/tex]

[tex]4x+8=x^2+4x[/tex]

Subtract 4x from both sides.

[tex]8=x^2[/tex]

Taking square root on both sides.

[tex]\pm\sqrt{8}=x[/tex]

[tex]\pm2\sqrt{2}=x[/tex]

Therefore, solutions of the given equation are [tex]x=2\sqrt{2}[/tex] and [tex]x=2\sqrt{2}[/tex].

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