Ta có: \(f\left(x_1\right)=g\left(x_1\right)\)

=>\(x_1^2+a\cdot x_1+b=x_1^2+c\cdot x_1+d\)

=>\(\left(a-c\right)\cdot x_1+b-d=0\) (1)

Ta có: \(f\left(x_2\right)=g\left(x_2\right)\)

=>\(x_2^2+a\cdot x_2+b=x_2^2+c\cdot x_2+d\)

=>\(\left(a-c\right)\cdot x_2+b-d=0\left(2\right)\)

Từ (1),(2) suy ra a-c=0

=>a=c

=>\(f\left(x\right)=x^2+a\cdot x+b;g\left(x\right)=x^2+a\cdot x+d\)

f(x)=g(x)

=>b=d

Đặt f(x)=0

=>\(x^2-3x-5=0\)

=>\(x^2-3x+\frac94-\frac{29}{4}=0\)

=>\(\left(x-\frac32\right)^2=\frac{29}{4}\)

=>\(\left[\begin{array}{l}x-\frac32=\frac{\sqrt{29}}{2}\\ x-\frac32=-\frac{\sqrt{29}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{29}+3}{2}\\ x=\frac{-\sqrt{29}+3}{2}\end{array}\right.\)

\(g\left(\frac{\sqrt{29}+3}{2}\right)=\left(\frac{3+\sqrt{29}}{2}\right)^2-4=\frac{\left(3+\sqrt{29}\right)^2-16}{4}\)

\(=\frac{9+29+6\sqrt{29}-16}{4}=\frac{22+6\sqrt{29}}{4}=\frac{11+3\sqrt{29}}{2}\)

\(g\left(\frac{-\sqrt{29}+3}{2}\right)=\left(\frac{-\sqrt{29}+3}{2}\right)^2-4\)

\(=\frac{29+9-6\sqrt{29}}{4}-4=\frac{38-6\sqrt{29}-16}{4}=\frac{22-6\sqrt{29}}{4}=\frac{11-3\sqrt{29}}{2}\)

\(g\left(x_1\right)\cdot g\left(x_2\right)\)

\(=\frac{11+3\sqrt{29}}{2}\cdot\frac{11-3\sqrt{29}}{2}=\frac{121-9\cdot29}{4}=\frac{-140}{4}=-35\)

Cho x các giá trị bất kì x 1 ,   x 2 sao cho  x 1   <   x 2

= >   x 1   -   x 2   <   0

Ta có:

f x 1 = 3 x 1 ; f x 2 = 3 x 2 ⇒ f x 1 − f x 2 = 3 x 1 − 3 x 2 = 3 x 1 − x 2 < 0 ⇒ f x 1 < f x 2

Vậy với   x 1   <   x 2 ta được f ( x 1 )   <   f ( x 2 )  nên hàm số y = 3x đồng biến trên tập hợp số thực R.