Answer to Question #94127 in Physics for Roaia

Question

If f(x)= x^2+bx+c x=2 is one of the roots of f(x)=0 so f(2) =

Answer 1

Let us consider a function $f(x)=x^2+bx+c$.

Solutions of the equation $f(x)=0$ satisfy the Vieta's formulas

$$x_1+x_2=-b, \quad x_1x_2=c$$

In our case

$$2+x_2=-b, \quad 2x_2=c$$

From these equations we get

$$2b+c=-4$$

Finally

$$f(2)=2^2+2b+c=4+2b+c=4-4=0$$

$$f(2)=0$$