The function f is defined by f(x) = ax^2 + bx + c, where a, b and c are constants. The equation f(x) = 0 has solutions x = 1 and x = -7. If a is an integer greater than 1, then a + b is greater than or equal to n for every such function f. What is the greatest possible value of n?

Step-by-step solution

1. 1

   Turn the two roots into a factored form. A quadratic is determined by its roots up to one scale factor. Since $f(1)=0$ and $f(-7)=0$, the factors $(x-1)$ and $(x+7)$ both divide $f$, so

   $f(x)=a(x-1)(x+7)$

   for some constant $a$ — and that $a$ is the same leading coefficient as in $ax^{2}+bx+c$, because the product $(x-1)(x+7)$ is monic.

2. 2

   Expand to read off $b$ in terms of $a$.

   $f(x)=a\left(x^{2}+6x-7\right)=ax^{2}+6ax-7a,$

   so matching coefficients gives $b=6a$ and $c=-7a$. The single unknown $a$ now controls everything.

3. 3

   Express the quantity being bounded.

   $a+b=a+6a=7a.$

   So the question "how large can $n$ be if $a+b\ge n$ always holds?" is really "how small can $7a$ get?"

4. 4

   Minimise over the allowed values of $a$. The constraint is that $a$ is an integer greater than 1, so the possible values are $a=2,3,4,\dots$ and the smallest is $a=2$. (Note $a\ne 0$ is automatic here — a quadratic needs a nonzero leading coefficient.) The smallest value of $a+b$ is therefore

   $7(2)=14.$

5. 5

   Convert the minimum into the answer. The inequality $a+b\ge n$ must hold for every allowed $f$, so $n$ can be at most the smallest value $a+b$ ever takes. That smallest value, $14$, is attained (at $a=2$, giving $f(x)=2x^{2}+12x-14$), so $n=14$ works and nothing larger does:

   $n_{\max}=14.$