Question

how can I find all complex solutions of

(4/x^2-3x)-(1/x^2-9)=0

Answer 1

\[\frac{4}{x^2 - 3x} - \frac{1}{x^2 - 9} = 0\]

Answer 2

\[\frac{4}{x^2 - 3x} = \frac{1}{x^2 - 9}\]

Answer 3

\[4(x^2 - 9) = 1(x^2 - 3x)\]
\[4x^2 - 36 = x^2 - 3x\]
\[3x^2 + 3x - 36 = 0\]
\[x^2 + x - 12 = 0\]
\[(x + 4)(x - 3) = 0\]
\[x = {-4,3}\]

I don't think this has any complex solutions.

Answer 4

@Diana0920

Answer 5

@Hero thanks (: