Is it possible that $mx^2-x+1=0$ has two negative roots?

Question

If the expression $(mx-1+\frac1x)$ is non-negative for all positive real $x$, then find the minimum value of $m$.

Given, $\frac{mx^2-x+1}x\ge0$ for all positive real $x$.

$\implies mx^2-x+1\ge0$ for all positive real $x$.

If $m\gt0$ and discriminant$\le0\implies1-4m\le0$, I am getting minimum value of $m$ as $1/4$, which is the correct answer.

But I wonder what if $m\gt0$, Discriminant$\ge0$ and both roots are negative? (Here, the quadratic is positive for positive $x$). Do we need to consider this? If yes, how to reach the answer in this case?

Answer 1

The question in your title does not really correspond with your text. The point of the text question is that you are looking for cases when the quadratic has no (real) roots, or just a single root. But I will answer the question in the title.

Think of the graph of $y=mx^2-x+1$. It cuts the $y$ axis at $y=1$ which is of course positive. If $m$ is negative, then the parabola is concave downwards and will have two roots, one positive and one negative. (Draw the graph!)

If $m$ is positive then the parabola is concave upwards and the vertex is at $x=1/(2m)$ which is positive. If there are two roots, then one will be between $x=0$ and $x=1/(2m)$, therefore positive; and the other will be greater than $x=1/(2m)$, therefore also positive. (Draw the graph!)

Answer 2

Assume $m \ne0.$ The equation is equivalent to $$x^2 -\frac{x}{m}+\frac{1}{m}=0.$$ The product of the roots is $1/m$. If both roots are negative, $m$ must be positive. The sum of the roots is $1/m.$ But then if both roots are negative, $m$ must be negative. Contradiction.