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 ?

There is no quadratic $x^2+ax+a$ with two real roots both greater than 5. — almagest, Apr 17 '16 at 12:39

almagest · Answer 1 · 2016-04-17T13:13:34.903

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.

edited Apr 17 '16 at 13:13

answered Apr 17 '16 at 12:52

almagest

score 0 · Answer 2 · answered Apr 17 '16 at 15:48

0

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

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

answered Apr 17 '16 at 15:48

Mick

2 Answers2