# If f(x) = ax^2 + bx + 2 and g(x) = bx^2 + ax + 1, is fog(x) = gof(x) for any value of x.

*print*Print*list*Cite

### 1 Answer

The function f(x) = ax^2 + bx + 2 and g(x) = bx^2 + ax + 1.

fog(x)

= f(g(x))

= f(bx^2 + ax + 1)

= a(bx^2 + ax + 1)^2 + b*(bx^2 + ax + 1) + 2

gof(x)

= g(f(x))

= g(ax^2 + bx + 2)

= b*(ax^2 + bx + 2)^2 + a*(ax^2 + bx + 2) + 1

gof(x) = fog(x)

=> a(bx^2 + ax + 1)^2 + b*(bx^2 + ax + 1) + 2 = b*(ax^2 + bx + 2)^2 + a*(ax^2 + bx + 2) + 1

=> a*b^2*x^4+2*a^2*b*x^3+(b^2+2*a*b+a^3)*x^2+(a*b+2*a^2)*x+b+a+2 = a^2*b*x^4+2*a*b^2*x^3+(b^3+4*a*b+a^2)*x^2+(4*b^2+a*b)*x+4*b+2*a+1

This is a very complex equation of the fourth order. It surely has a solution for some values of coefficients a and b.

**It is not possible to determine the required values of x with unknown coefficients a and b.**