+0  
 
0
111
1
avatar+123 

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! 

 Oct 19, 2018
edited by hearts123  Oct 20, 2018
edited by hearts123  Oct 25, 2018
 #1
avatar+98044 
+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

 

 

cool cool cool

 Oct 20, 2018
edited by CPhill  Oct 20, 2018

34 Online Users

avatar
avatar
avatar