Find $m$ so that equation has two roots inside given interval

Question

We are given the following equation:

$$x^2 - (2m - 5)x + 3m - 1 = 0$$

We have to find $m \in \mathbb{R}$ so that the given equation has two distinct real roots in $[1, 2]$.

In order for the equation to have two distinct real roots, the discriminant has to be greater than 0. This is the first condition.

The second condition I found is that $f(1) \cdot f(2) < 0$ ($f$ is a function denoting the left part of the equation above), that is because $f$ must intersect the $X$ axis between $1$ and $2$.

However, these two conditions are not enough, I need one more.

Thank you in advance for your help!

Answer 1

Hint:

If $x_1$ and $x_2$ are the two roots in the interval $[1,2]$ than: $$ 2\le x_1+x_2\le 4 \qquad \mbox{and}\qquad 1\le x_1x_2\le 4 $$

so:

$$ \begin{cases} 2m-5\ge2\\ 2m-5\le4\\ 3m-1\ge 1\\ 3m-1\le 4 \end{cases} $$

what you can find from this?

Answer 2

HINT

For $f(x)=ax^2 +bx +c$ to have two distinct real roots in [1,2] the conditions are

1. discriminant greater than 0

2. $\frac {-b} {2a} \in [1,2]$

3. $f(1) \cdot f(\frac {-b} {2a}) \le 0$ and $f(2) \cdot f(\frac {-b} {2a}) \le 0$ and $f(\frac {-b} {2a}) \ne 0$

Answer 3

$$\Delta = (2m-5)^2-4(3m-1)= 4(m-4)^2-35$$

For $\Delta \ge 0$, $$m \ge 4+\frac{\sqrt{35}}{2} \quad \text{or} \quad m \le 4-\frac{\sqrt{35}}{2}$$

$$2m-5 \ge 3+\sqrt{35} \quad \text{or} \quad 2m-5 \le 3-\sqrt{35}$$

$$2m-5 > 3+5 \quad \text{or} \quad 2m-5 < 3-5$$

$$\frac{\alpha+\beta}{2} > 4 \quad \text{or} \quad \frac{\alpha+\beta}{2} < -1$$ which beyond $[1,2]$

It's impossible to have all real roots in $[1,2]$