Question

$$x^{2}-bx+9=0$$

Solve for b

$b=x+\frac{9}{x}$
$x\neq 0$

Show Solution

Solve for x (complex solution)

$x=\frac{\sqrt{b^{2}-36}+b}{2}$
$x=\frac{-\sqrt{b^{2}-36}+b}{2}$

Solve for x

$x=\frac{\sqrt{b^{2}-36}+b}{2}$
$x=\frac{-\sqrt{b^{2}-36}+b}{2}\text{, }|b|\geq 6$