The values of $x_1 \cdot x_2$ from $x^{2}-2mx+2m^{2}-2m=0$

Question

I have the following equation: $x^{2}-2mx+2m^{2}-2m=0$ with the real roots $x_1,x_2$

I need to find the values of the product of $x_1\cdot x_2$.The right answer is $[-\frac{1}{2},4]$

My try: To have real roots, the discriminant of the equation is positive so I got $m\in[0,2]$.Now, $x_1\cdot x_2=2m^{2}-2m=2m(m-1)$

$0\leq m\leq 2$ so $0\leq 2m\leq 4$

$-1\leq m-1\leq 1$ so $0\leq 2m(m-1)\leq 4$ but it's wrong.Where's my mistake?How to approach this exercise?

Answer 1

Your approach is correct, the only problem is finding the range of $2m(m-1)$ on the interval $[0,2]$

Finding the maximum and the minimum of your function is done by taking derivative and also evaluating your function at the endpoints and comparing those values. You will get $-1/2$ and $4$ for your minimum and maximum respectively.

Answer 2

Complete squares to get $2m^2-2m=2(m-\frac{1}{2})^2-\frac{1}{2}$. Now note that $m\in[0,2]$ then $m-1/2\in[-\frac{1}{2},\frac{3}{2}]$. Then $(m-\frac{1}{2})^2\in [0,\frac{9}{4}]$. So $$ 2(m-1/2)^2-1/2\in[2(0)-\frac{1}{2},2(\frac{9}{4})-\frac{1}{2}]=[-\frac{1}{2},4]$$