Given the following equation:
[tex]\frac{x}{x-4}-\frac{4}{x}=\frac{3}{x-4}[/tex]
First, we will identify the zeros of the denominator
So, the zeros are: x = {0,4}
Second, multiply the equation by x(x-4) to eliminate the denominators
[tex]x(x-4)*(\frac{x}{x-4}-\frac{4}{x})=x(x-4)*\frac{3}{x-4}[/tex]
Simplify the equation:
[tex]x^2-4(x-4)=3x[/tex]
Expand the equation and combine the like terms:
[tex]\begin{gathered} x^2-4x+16=3x \\ x^2-7x+16=0 \end{gathered}[/tex]
The last quadratic equation will be solved using the quadratic rule:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Substitute a = 1, b = -7, c = 16
[tex]\begin{gathered} x=\frac{7\pm\sqrt{(-7)^2-4(1)(16)}}{2(1)} \\ \\ x=\frac{7\pm\sqrt{-15}}{2}=\frac{7\pm i\sqrt{15}}{2} \\ \\ x=\lbrace\frac{7+i\sqrt{15}}{2};\frac{7-i\sqrt{15}}{2}\rbrace \end{gathered}[/tex]
So, the answer will be:
[tex]x=\lbrace\frac{7+i\sqrt{15}}{2};\frac{7-i\sqrt{15}}{2}\rbrace[/tex]
