Let f(x) be a quadratic polynomial with leading coefficient 1 such that f(0)=p, p ≠ 0, and f(1)=1/3. If the equations f(x)=0 and f ° f ° f ° f(x)=0 have a common real root, then f(-3) is equal to ________________.

Solution

Maths sir · 5 steps

Start from the common root and apply f again and again: the root turns into 0, then into p.

1. Set up f. Write f(x) = x² + bx + p. From f(1) = 1/3 we get 1 + b + p = 1/3, so b = -2/3 - p.
2. Follow the root. Let α be the common root, so f(α) = 0. Then f(f(α)) = f(0) = p, the third application gives f(p), and the fourth gives f(f(p)). The condition f ° f ° f ° f(α) = 0 says f(f(p)) = 0, so f(p) is itself a root of f.

   f(p) = p² + bp + p = p(p + b + 1) = p(p - 2/3 - p + 1) = p/3

3. Find p. So f(p/3) = 0: p²/9 + bp/3 + p = 0. Divide by p ≠ 0 to get p + 3b + 9 = 0. Put in b = -2/3 - p: p - 2 - 3p + 9 = 0, so p = 7/2 and b = -25/6.
4. Check real roots. f(x) = x² - 25/6x + 7/2 has discriminant 625/36 - 14 = 121/36 > 0, with roots 3 and 7/6. The root 7/6 = p/3 is real, as required.
5. So. f(-3) = 9 + 25/6 × 3 + 7/2 = 9 + 25/2 + 7/2 = 9 + 16 = 25.

The idea to keep. When f and a composition of f share a root, put the root into the composition. f(α) = 0 turns the chain into f(0), f(f(0)), …, which are just numbers.

Answer from the NTA final answer key. Solution by PrepWiser.