# Location of roots of a Quadratic Equation

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.

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

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 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. It seems that you missed that possibility.

• 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. – General Kenobi Jul 11 at 11:44
• Why is that? Can you justify it? – José Carlos Santos Jul 11 at 11:48