5. Variant 1.

On the coordinate plane, the graphs of three reduced quadratic trinomials are drawn, intersecting the y-axis at points $-15, -6, -27$ respectively. For each trinomial, the coefficient of $x$ is a natural number, and the larger root is a prime number. Find the sum of all roots of these trinomials.
Answer: -9.

## Solution.

The parabola intersects the $O y$ axis at a point whose ordinate is equal to the free term. Therefore, our trinomials have the form $x^{2}+a x-15, x^{2}+b x-6, x^{2}+c x-27$. Let's denote their larger roots by $p, q, r$ respectively. By Vieta's theorem, the second root of the first trinomial is $-a-p$, then $p(-a-p)=-15$ or $p(p+a)=15=3 \cdot 5$. Similarly, $q(q+b)=6=2 \cdot 3, r(r+c)=27=3 \cdot 9$.

Since $p+a>p, q+b>q, r+c>c$ and $p, q, r$ are prime, then $p=3, q=2, r=3$ and $p+a=5$, $q+b=3, r+c=9$. Therefore, the sum of the roots is $3+(-5)+2+(-3)+3+(-9)=-9$.