+0  
 
0
17
1
avatar+1612 

For what real values of $k$ does the quadratic $12x^2 + kx + 27 = 6x^2 + 12x + 18$ have nonreal roots? Enter your answer as an interval.

 Jan 11, 2024
 #1
avatar+128732 
+1

Simplify  as

 

6x^2 + ( k - 12)x + 9  = 0

 

We have non-real roots when the discriminant > 0

 

So

 

(k  -12)^2  - 4(6)(9) < 0

 

(k -12)^2  -  216 <  0

 

(k -12)^2  < 216    take both roots

 

k - 12 < sqrt (216)                and        k - 12 >  -sqrt (216)

 

k < 12 + 6sqrt (6)                                 k > 12 - 6 sqrt (6) 

 

k  on this interval →  ( 12 -6sqrt(6)  , 12 + 6sqrt (6) )                                     

 

 

 

cool cool cool

 Jan 11, 2024

2 Online Users

avatar
avatar