TITLE: For what values of m are the roots of $x^2 +2x+3 = m(2x+1)$ real and positive
QUESTION [6 upvotes]: I am only able to show that to be real, $m <-1$ or $m\geq2$
Don't know how to finish solution
Answer is $2 \leq m < 3$
So far:
After expanding and factorising, 
$x^2 + 2(1-m)x + (3-m) = 0  $
Roots are real if discriminant $\ge 0 $ 
i.e. $4(m-2)(m+1)>=0  $
Therefore $m ≤ -1 \text{ or } m \ge 2$ for roots to be real
However I don't know how to find m for roots to be positive

REPLY [3 votes]: For the equation to have positive and real roots, two extra conditions are needed
$$2m-2>0$$ $$3-m>0$$ 
$$\therefore 1<m<3$$
Taking the intersection of the intervals found, we will get $$\therefore 2 \leq m <3$$
:)