Find $m$ if the equation $x^2+m^2-2mx+3x-5m+3=0$ has roots numerically equal but opposite in sign.

First rewrite the equation as $x^2+(-2m+3)x+(m^2-5m+3)=0$. If the roots are equal in magnitude and opposite in sign, their sum is 0, so the coefficient of $x$ must be 0. Hence $-2m+3=0$, giving $m=\frac32$, which is not among the options; therefore we must check the intended textbook condition carefully. Using the standard rule from the chapter, opposite roots imply $b=0$, so the required value should satisfy $-2m+3=0$. That gives $m=\frac32$. Since this value is absent, the printed item appears inconsistent; among the given options none is correct.