Problem - 2602
Consider the equation $x^2 +(m-2)x + \frac{1}{2}m-3=0$.
(1) Show that this equation always have two distinct real roots
(2) Let $x_1$ and $x_2$ be its roots. If $x_1+x_2=m+1$, what is the value of $m$?
(1) The discriminant of this equation equals
$$(m-2)^2-4\cdot\Big(\frac{1}{2}m-3\Big)=m^2 - 4m+4 - 2m +6=m^2-6m+10=(m-3)^2+1 > 0$$
Hence, it must always have two distinct roots.
(2) By Vieta's theorem, $x_1+x_2=-(m-2)$. Therefore
$$-(m-2)=m+1 \implies m=\boxed{\frac{1}{2}}$$