+0  
 
0
196
1
avatar

Find the set of values of k for which this equation has two real roots.
5x^2 + 4k*x + k = x^2 + 3x + 2

 Jun 28, 2022
 #1
avatar+128448 
+1

Rearrange as

 

4x^2 + (4k - 3)x  + k - 2  =  0

 

This will have two real roots when the discriminant  >   0

 

So

 

(4k - 3)^2  -  4 (4) (k - 2)  >  0

 

16k^2 - 24k + 9  - 16k + 32 >  0

 

16k^2 - 40k + 41  >  0

 

See the graph here (in x)  :  https://www.desmos.com/calculator/bwva5lob60

 

This parabola  lies wholly above the x axis so any real  value of  k will give two real solutions

 

 

cool cool cool

 Jun 28, 2022

7 Online Users

avatar
avatar