The function f is defined by f(x) = ax² + 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 ≥ n for every such f. What is the greatest possible value of n?

Step-by-step solution

1. 1

   Turn the roots into a factored form. A quadratic with roots $1$ and $-7$ must be a constant multiple of $(x-1)(x+7)$, and that constant is the leading coefficient:

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

   This is the key move: knowing both roots leaves exactly one free parameter, $a$, instead of three.

2. 2

   Expand to read off $b$.

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

   so $b=6a$ and $c=-7a$. Note $b$ is not free — it is locked to $a$.

3. 3

   Express the quantity being bounded.

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

   The question is therefore about how small $7a$ can be.

4. 4

   Minimise over the allowed values of $a$. The constraint "$a$ is an integer greater than $1$" means $a\in\{2,3,4,\dots\}$, so the smallest is $a=2$, giving $a+b=14$. For every larger integer $a$ the value $7a$ is bigger, so

   $a+b\ge 14\quad\text{for every such }f.$

5. 5

   Identify the greatest valid $n$. Any $n\le 14$ makes the statement "$a+b\ge n$ for every such $f$" true, but $n=15$ fails at $a=2$, where $a+b=14<15$. So the greatest possible value is $n=14$. Check: $a=2$ gives $f(x)=2x^{2}+12x-14$, whose roots are indeed $1$ and $-7$, with $a+b=2+12=14$.