Given a quadratic equation: x^2 - sqrt(5a^2 - 26a - 8) * x - (a^2 - 4a + 9) = 0

It has 2 integer roots. If a is an integer as well, what is the value of a?

edit: thank you, I really appreciate your help! :)

edit 2: thank you for taking the time to correct it!

hearts123 Oct 19, 2018

#1**+3 **

To have two roots.....the discriminant must be a perfect square > 0.....so...

5a^2 - 26a - 8 + 4 (1)(a^2 - 4a + 9) simplify

9a^2 - 42a + 28

Graphing this...it will be > 0 and a perfect square when the integer value for a = 6

So the polynomial is

x^2 - 4x - 21 factor ( x - 7) ( x + 3) and the roots are - 3 and 7

CORRECTED

CPhill Oct 20, 2018