Question

find all real and complex solutions using the quadratic formula: x=-9/x+3

Answer 1

\(\bf x=-\cfrac{9}{x+3} ?\)

Answer 2

yes

Answer 3

can it be simplified further?

Answer 4

\(\large\color{black}{ \displaystyle x=\frac{-9}{x+3} }\)

multiply both sides times \(\large\color{black}{ \displaystyle x+3 }\)

\(\large\color{black}{ \displaystyle x\color{blue}{\times (x+3)}=\frac{-9}{x+3} \color{blue}{\times (x+3)} }\)

and you get: \(\large\color{black}{ \displaystyle x^2+3x=-9 }\)

\(\large\color{black}{ \displaystyle ( }\) the (\(\normalsize\color{black}{ \displaystyle x+3 }\))'s canceled on the right side \(\large\color{black}{ \displaystyle ) }\)

Answer 5

now, you can add 9 to both sides and use the quadratic formula, just as they want you to.

Answer 6

okay

Answer 7

\(\bf x=-\cfrac{9}{x+3}\implies x(x+3)=-9\implies x^2+3x=-9
\\ \quad \\
x^2+3x+9=0\\

{\color{blue}{ 1}}x^2{\color{red}{ +3}}x{\color{green}{ +9}}=0
\qquad \qquad
x= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\)

as shown already by SolomonZelman above