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.

  • $\begingroup$ Your intersection is not correct. $\endgroup$ – Akshat Sharma Jul 11 at 11:38
  • $\begingroup$ I know that, but I can't understand why. $\endgroup$ – General Kenobi Jul 11 at 11:47
  • $\begingroup$ I think $f(3) \lt 0$ is not correct since there would be two roots , one in $(2,3)$ and other $\gt 3$ $\endgroup$ – user710290 Jul 11 at 11:50
  • 1
    $\begingroup$ You are missing the case where the axis of the parabola is less than $x\lt 2$. Consider $f(2)f(3)\lt 0$. $\endgroup$ – mathlove Jul 11 at 11:56

There are two possible cases.

Case 1. $f(2) \lt 0$ and $f(3) \gt 0$.


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

enter image description here

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Yes, there is exatly one root in $(2,3)$ if $f(2)>0>f(3)$. But it is also true that there is exatly one root in $(2,3)$ if $f(2)<0<f(3)$. It seems that you missed that possibility.

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  • $\begingroup$ Actually, there would be a root in $f(2)<0<f(3)$ if the leading coefficient of the equation would be negative and not positive. $\endgroup$ – General Kenobi Jul 11 at 11:44
  • $\begingroup$ Why is that? Can you justify it? $\endgroup$ – José Carlos Santos Jul 11 at 11:48

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