Quadratic Equation $9x^2-\text{ln} c = bx$

Question

How to determine the roots ofthe quadratic equation $9x^2 -\text{ln} c= bx$, where $1<c<3$,

$(1)$ real and distinct roots

$(2)$ real and equal roots

$(3)$ no real roots.

Answer 1

Recall that $Ax^2+Bx+C=0$ has real and distinct real roots if and only if

$$B^2-4AC>0$$

therefore for $9x^2 -\ln c= bx \iff 9x^2 -bx -\ln c= 0$ we need

$$b^2+36\ln c>0$$

Answer 2

The discriminant is $b^2+36\ln{c}$. Note that since $c\in(1,3)$, $\ln c$ starts from (not including) $0$ and goes upto a little over 1 (and $\ln$ is strictly increasing). This means the discriminant is surely positive ($b^2$ of course is always positive) and so the roots are real. For the roots to be equal, $b^2+36\ln c = 0$ but this is impossible since both $b^2$ and $36\ln c$ are strictly positive.

$$\boxed{\text{The roots are real and distinct. }}$$