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avatar+41 

given the equation: (a+1)x^2 - (a^2 + 1)x + 2a^3 - 6 = 0

what's the value for a (which is an integer) that makes the equation have all integer roots? 

 

edit: thank you sm :)

hearts123  Oct 25, 2018
edited by hearts123  Oct 27, 2018
 #1
avatar+90995 
+3

 (a+1)x^2 - (a^2 + 1)x + 2a^3 - 6 = 0

 

By inspection...let  a  =  0   and we have

 

x^2  - x   - 6  =  0     factoring, we get

 

(x - 3) ( x + 2)  = 0

 

And the two  roots  are  x  = 3  and x  = -2

 

 

 

cool cool cool

CPhill  Oct 25, 2018
 #2
avatar+90995 
+1

P.S.....there may be other solutions for  "a"....I don't  know.....perhaps someone else knows another way to  obtain all of them,  if there are more...!!!!

 

 

cool cool cool

CPhill  Oct 25, 2018
 #3
avatar+27128 
+3

If a = 1 there are solutions x = -1 and x = 2

 

If a = -1 the quadratic reduces to a linear equation with solution x = -4

.

Alan  Oct 26, 2018
 #4
avatar+90995 
+1

Thanks, Alan !!!!

 

 

cool cool cool

CPhill  Oct 26, 2018

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