Solutions from an equation

By Vieta, we immediately have $x_1+x_2=2(m-1)$ and $x_1 x_2=2m^2-3m+1=(2m-1)(m-1)$ ; so $A=\left| (2m+1)(m-1) \right|$

We also need the original equation to have real roots; hence we need $(m-1)^2>2m^2-3m+1$

Expanding and rearranging, this condition is $m(m-1)<0$

ie $0<m<1$ .

In this range, $m-1<0$ and $2m+1>0$ so that $A=(2m+1)(1-m)$

Either by differentiating or completing the square, we find this has a maximum at $m=\frac14$ (which is in the required range), where $A=\boxed{\frac98}$ .

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

$\Rightarrow \Delta = [-2(m-1)]^2 - 4(2m^2-3m+1) = 4m^2-8m+4 - (8m^2 - 12m+4) = -4m^2 +4m = 4m(1-m)$

In order that the equation has 2 distinct real solutions, $\Delta > 0 \Leftrightarrow 4m(1-m) > 0$

$\Leftrightarrow 0 < m < 1$

Using Vieta's formula for Quadratics , we have:

$\left\{\begin{matrix} x_1+x_2=2(m-1)\\ x_1x_2= 2m^2-3m+1 \end{matrix}\right.$

Substituting these results into $A$ , we have:

$A=|(x_1+x_2)+x_1x_2| = |2(m-1) + 2m^2 - 3m+1| = |2m^2-m-1| = |(m-1)(2m+1)|$

Using $0 \leq m \leq 1$ , we can deduce that $\left\{\begin{matrix} m-1 \leq 0\\ 2m+1 \geq 0 \end{matrix}\right.$

Therefore, $(m-1)(2m+1) \leq 0$

Hence, $A=|(m-1)(2m+1)| = -(m-1)(2m+1) = -2m^2 +m +1$

$\Leftrightarrow A= -2(m^2-2.m.\frac{1}{4}+\frac{1}{16}) +\frac{1}{8} + 1 = -2(m-\frac{1}{4})^2 + \frac{9}{8} \leq\frac{9}{8}$

The inequality $A \leq \frac{9}{8}$ occurs only when $m-\frac{1}{4} = 0 \Leftrightarrow m=\frac{1}{4}$ (satisfies the condition $0 < m < 1$ , which is necessary for the first equation to have 2 distinct real solutions)

Therefore, the maximum value of $A$ is $\boxed{\frac{9}{8}}$

For the equation to have two real roots, the discriminant

$\begin{aligned} (2(m-1))^2 - 4(2m^2-3m+1) & > 0 \\ 4m^2 - 8m + 4 - 8m^2 + 12m - 4 & > 0 \\ - 4m^2 + 4m & > 0 \\ \implies m(m-1) & < 0 \end{aligned}$

This implies that the equation has two real roots, when $0< m < 1$ . Then we have:

$\begin{aligned} |\blue{x_1+x_2}+\red{x_1x_2}| & = |\blue{2(m-1)} + \red{2m^2-3m+1}| & \small \blue{\text{By Vieta's formula}} \\ & = |2m^2 - m - 1| \\ & = \left| 2\left(m - \frac 14\right)^2 - \frac 98 \right| \\ \implies \max |x_1+x_2 + x_1x_2| & = \left|-\frac 98 \right| = \boxed{1.125} & \small \blue{\text{when }m = \frac 14} \end{aligned}$

---

Reference: Vieta's formula