Question

If \[ f(f(0))=0, \] where \[ f(x)=x^2+ax+b,\qquad b\neq0, \] then \(a+b=\)

• A. \(2\)
• B. \(1\)
• C. \(-1\)
• D. \(-2\)

Hint:

Whenever composite functions such as \(f(f(0))\) are given, first compute the inner function carefully and then substitute into the outer function.

Answer 1

The Correct Option is C

Step 1: Find $f(0)$.
Since $f(x)=x^2+ax+b$, substituting $x=0$ gives $f(0)=b$.
Step 2: Apply the nesting condition.
We are told $f(f(0))=0$, that is $f(b)=0$.
Step 3: Compute $f(b)$.
$f(b)=b^2+ab+b$.
Step 4: Set it to zero and factor.
$b^2+ab+b=0$ gives $b(b+a+1)=0$.
Step 5: Use the given restriction.
Since $b\neq 0$, we must have $a+b+1=0$.
Step 6: Solve for $a+b$.
Therefore $a+b=-1$, which is option (3).
\[ \boxed{-1} \]