Problem 8 · A parabola and the equation f(x)·f(x+k) = 0
Discriminant: when a product of two parabolas has exactly two roots.
Integer answer, at most 4 digitsIf $f(x) = x^{2} - bx + 9$, with $b > 0$, what value of $b$ makes the equation $f(x)\cdot f(x+k) = 0$ have exactly two distinct real solutions whenever $k \neq 0$?
Reasoned solution
Key idea: $f(x)\cdot f(x+k) = 0$ means $f(x)=0$ or $f(x+k)=0$. The solutions of the latter are those of the former shifted by $k$ units.
If $f$ has two distinct roots $r \neq t$ (positive discriminant), the product has solutions $\{r,\ t,\ r-k,\ t-k\}$: for a generic $k \neq 0$ these are four distinct values — too many.
If $f$ has no real roots (negative discriminant), neither does the product: no solutions.
The only way to get exactly two for every $k \neq 0$ is for $f$ to have a double root:
Check: $f(x) = (x-3)^{2}$, and $f(x)\,f(x+k) = 0$ has solutions $x = 3$ and $x = 3-k$: exactly two distinct ones for any $k \neq 0$. ✓