Let f(x)=x2-6 x+8, then number of real roots of the equation f(f(x))=0 is

Question

Let f(x)=x2-6 x+8, then number of real roots of the equation f(f(x))=0 is (A) 4 (B) 2 (C) 1 (D) 0

Tardigrade Admin · Accepted Answer

f(x)=x2-6 x+8 f ( f ( x ))=0 ⇒ f ( x )=2 or f ( x )=4 ⇒ f ( f ( x ))=0 has four distinct real roots.

Q. Let $f (x) = x^{2} - 6 x + 8$ , then number of real roots of the equation $f (f (x)) = 0$ is