Let f(x)=(a−3)x2+12x+(a+6) to quadratic always be positive a−3>0 ⇒a>3 and D=(12)2−4(a−3)(a+6)<0 =144−4(a2+3a−18)<0 ⇒a2+3a−54>0 ⇒a2+9a−6a−54>0 ⇒(a+9)(a−6)>0
a∈(−∞,−9)∪(6,∞) Combine with a>3 ⇒ We take a∈(6,∞) Least positive integeral value α=7,l=6 put these values (α−3)x2+12x+(l+2) =4x2+12x+8=0 ⇒x2+3x+2=0 ⇒(x+1)(x+2)=0 Hence, x=−1,−2c
