Given f(x)=(x2+3x)(x2+3x+2)
Let u=x2+3x, so:
f(x)=u(u+2)=u2+2u
f(x)+1=u2+2u+1=(u+1)2=(x2+3x+1)2
So the equation f(x)+1=9701 becomes:
(x2+3x+1)2=9701
∣x2+3x+1∣=9701
Case 1: x2+3x+1=9701
x2+3x−9700=0
Δ=9+4(9700)=38809=1972>0
This gives two real roots.
Case 2: x2+3x+1=−9701
x2+3x+9702=0
Δ=9−4(9702)=−38799<0
No real roots.
All real roots come from Case 1: x2+3x−9700=0
Using Vieta's formulas, sum of roots =−ab=−13=−3