Try this option:
1. if 8*x²+b*x+3=8*x²+p*x+q*x+3, then ⇒ b=p+q.
2. p*q = max_value (b²/2), if p=q=0.5*b, and p*q→-oo, if p>0 and q<0 or p>0 q<0.
3. example:
given 8x²+10x+3, the student rewrites it as a) 8x²+5x+5x+3 (5*5=25-max value); b) 8x²+0.01x+9.99x+3 (9.99*0.01=0.0999→0); c) 8x²-20x+30x+3 (p*q=-600).
answer: (-oo;0.5b²)
