Ta có : \(\left\{{}\begin{matrix}f\left(a\right)=2a^2+b=0\\f\left(b\right)=b^2+ab+b=0\\2a^2=b^2+ab\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2a^2+b=0\\a+b=-1\\a^2-b^2=\left(a+b\right)\left(a-b\right)=ab-a^2=a\left(b-a\right)\end{matrix}\right.\)

\(\Rightarrow\left(a+b\right)\left(a-b\right)+a\left(a-b\right)=\left(a-b\right)\left(2a+b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=b\\a+b=-a=-1\end{matrix}\right.\)

TH1 : a = b .

\(\Rightarrow a=b=-\dfrac{1}{2}\)

TH2 : a = 1

\(\Rightarrow b=-2\)

\(f\left(a\right)=f\left(b\right)=x^2+ax+b=0\)

\(\Rightarrow ax+b=-x^2\)

\(\Rightarrow-\left(ax+b\right)=x^2\)

\(\Rightarrow-\left(ax+b\right)+ax+b=0\)

\(\Rightarrow-ax-b=ax+b=0\)

hay

\(\Rightarrow\left|-ax-b\right|=\left|ax+b\right|=0\)

\(\Rightarrow a=\frac{b}{x}\left(x\ne0\right)\)

Ta có \(f\left(a\right)=a^2+a^2+b=0\)

=> \(2a^2+b=0\)(1)

và \(f\left(b\right)=b^2+ab+b=0\)(2)

Từ (1) và (2) => \(2a^2+b=b^2+ab+b=0\)

=> \(2a^2-b^2-ab=b^2+b-b=0\)

=> \(2a^2-b^2-ab=b^2=0\)

=> \(2a^2-ab=b^2+b^2=0\)

=> \(2a^2-ab=2b^2=0\)

=> \(a\left(2a-b\right)=2b^2=0\)

=> \(\hept{\begin{cases}a\left(2a-b\right)=0\\2b^2=0\end{cases}}\)=> \(\hept{\begin{cases}a\left(2a-b\right)=0\left(1\right)\\b=0\end{cases}}\)

Thay b = 0 vào (1), ta có: a. 2a = 0

=> 2a² = 0

=> a² = 0 => a = 0.

Vậy a = b = 0.

Vì f(0)=4 => c=4

=> f(x)=ax^2+bx+4

Vì f(1)=3 => a+b+4=3 => a+b=-1(1)

f(-1)=7 => a-b+4=7 => a-b =3 (2)

Từ (1),(2) => a = 1; b=-2 

=> f(x)=x^2-2x+4

Với x=-2

\(f\left(-2\right)=-2a+b=0\)(1)

Với x=2

\(f\left(2\right)=2a+b=8\)(2)

Lấy (1)+(2) ta được

\(2b=8\Rightarrow b=4\Rightarrow a=2\)

Ta có:\(f\left(-2\right)=-2a+b\)

mà \(f\left(-2\right)-0\)

\(\Rightarrow-2a+b=0\)(1)

Lại có: \(f\left(2\right)=2a+b\)

mà \(f\left(2\right)=8\)

\(\Rightarrow2a+b=8\)(2)

Lấy (2) trừ (1) ta được:

\(4a=8\)

\(a=2\)

Thay a=2 vào (1) ta có b=4

Vậy a=2 ; b=4