We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
144
1
avatar+155 

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+100595 
+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

33 Online Users

avatar
avatar
avatar