How to solve specific parameters for a quadratic equation?

Question

x^2+ax+a

so that there are two different solutions x>5

First I set up that the discriminant is:

D > 0

Then using Vieta's formula:

a>25, a<10

But still, if I take 5 and 6 as solutions, I end up breaking my condition. How can I set up the conditions ?

Answer 1

The product of the roots of $x^2+ax+a=0$ is $a$ and their sum is $-a$, so the roots cannot both be positive.

Answer 2

For real roots, we need $a^2 – 4a \ge 0$

Solving that inequality, we have $a \le 0$ or $a \ge 4$