Question:
If
\[
f(f(0))=0,
\]
where
\[
f(x)=x^2+ax+b,\qquad b\neq0,
\]
then \(a+b=\)
If
\[
f(f(0))=0,
\]
where
\[
f(x)=x^2+ax+b,\qquad b\neq0,
\]
then \(a+b=\)
Show Hint
Whenever composite functions such as \(f(f(0))\) are given, first compute the inner function carefully and then substitute into the outer function.
Updated On: Jun 22, 2026
- \(2\)
- \(1\)
- \(-1\)
- \(-2\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Find \(f(0)\).
Given, \[ f(x)=x^2+ax+b \] Substituting \(x=0\), \[ f(0)=b \]
Step 2: Use the condition \(f(f(0))=0\).
Since \[ f(0)=b, \] we get \[ f(b)=0 \] Now, \[ f(b)=b^2+ab+b \] Thus, \[ b^2+ab+b=0 \]
Step 3: Factor the equation.
Taking \(b\) common, \[ b(b+a+1)=0 \] Given, \[ b\neq0, \] therefore, \[ a+b+1=0 \] Hence, \[ a+b=-1 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{-1} \] which corresponds to option (3).
Given, \[ f(x)=x^2+ax+b \] Substituting \(x=0\), \[ f(0)=b \]
Step 2: Use the condition \(f(f(0))=0\).
Since \[ f(0)=b, \] we get \[ f(b)=0 \] Now, \[ f(b)=b^2+ab+b \] Thus, \[ b^2+ab+b=0 \]
Step 3: Factor the equation.
Taking \(b\) common, \[ b(b+a+1)=0 \] Given, \[ b\neq0, \] therefore, \[ a+b+1=0 \] Hence, \[ a+b=-1 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{-1} \] which corresponds to option (3).
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