Let t=x≥0. The equation becomes (−2+3)∣t−3∣+(t2−6t)+(9−23)=0.
Since t2−6t+9=(t−3)2, we get (t−3)2+(3−2)∣t−3∣−23=0.
Let u=∣t−3∣≥0: u2+(3−2)u−23=0.
Discriminant =(3−2)2+83=7+43=(2+3)2.
u=2(2−3)±(2+3), giving u=2 or u=−3 (rejected).
∣t−3∣=2⇒t=5 or t=1, so x=25 or x=1.
α=1,β=25.
β/α+αβ=25+25=5+5=10.