Question:
For what values of $m\in\mathbb R$, the equation $2x^2-2(2m+1)x+m(m+1)=0$ has exactly one root in the interval $(2,3)$?
My Approach:
As the leading coefficient of the equation is positive, its graph would be an upwards opening parabola. This implies that $f(2)>0$ and $f(3)<0$
From $f(2)>0$, we get:
$$m^2-7m+4>0$$
which gives us
$$m\in(- \infty,\frac{7-\sqrt{33}}{2})\cup(\frac{7+\sqrt33}{2},\infty)$$
From $f(3)<0$, we get:
$$m\in(\frac{11-\sqrt73}{2},\frac{11+\sqrt73}{2})$$
Intersecting the above two intervals, I get:
$$m\in(- \infty,\frac{7-\sqrt{33}}{2})\cup(\frac{7+\sqrt33}{2},\frac{11+\sqrt73}{2})$$
However, the correct answer is:
$$m\in(\frac{7-\sqrt{33}}{2},\frac{11-\sqrt73}{2})\cup(\frac{7+\sqrt33}{2},\frac{11+\sqrt73}{2})$$$$
I can't figuire out how it is correct, even after finding out the values. Please help.
Best Answer
There are two possible cases.
Case 1. $f(2) \lt 0$ and $f(3) \gt 0$.
OR
Case 2. $f(2) \gt 0$ and $f(3) \lt 0$.
Also, I think the intersection of the two zones mentioned will be a continuous one.
Here is a diagramatic representation of the solution