\(\Delta=a^2-4\cdot1\cdot\left(b+2\right)=a^2-4b-8\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>\(a^2-4b-8>0\)
=>\(a^2>4b+8\)
Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=-a;x_1x_2=\frac{c}{a}=b+2\)
\(x_1^3-x_2^3=28\)
=>\(\left(x_1-x_2\right)\left(x_1^2+x_2\cdot x_1+x_2^2\right)=28\)
=>\(x_1^2+x_1x_2+x_2^2=7\)
=>\(\left(x_1+x_2\right)^2-x_1x_2=7\)
=>\(\left(-a\right)^2-\left(b+2\right)=7\)
=>\(a^2-b-2=7\)
=>\(a^2=b+9\)
\(x_1-x_2=4\)
=>\(\left(x_1-x_2\right)^2=16\)
=>\(\left(x_1+x_2\right)^2-4x_1x_2=16\)
=>\(\left(-a\right)^2-4\) (b+2)=16
=>b+9-4b-8=16
=>-3b+1=16
=>-3b=15
=>b=-5
\(a^2=b+9=\) -5+9=4
=>a=2 hoặc a=-2
