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January 12th, 2019, 04:33 AM  #1 
Newbie Joined: Jan 2019 From: Bosnia and Herzegovina Posts: 9 Thanks: 0  I've got solved assignment, but don't understand something
Assignment says: For which k is equation (k2)x²+2(k2)x+2=0 doesn't have real solutions? Solution: D<0 D=(2(k2))²4*2(k2) =4(k²4k+4)8(k2) =4k²16k+168k+16 =4k²24k+32<0 /:4 =k²6k+8 k1,2= 6±√4 / 2 k1 = 4 and k2 = 2 and solution is that k∈[2,4) Now I don't understand why 4 isn't element of k and 2 is? 
January 12th, 2019, 05:38 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,044 Thanks: 1627 
I do not agree with the given solution. discriminant must be strictly negative for the quadratic to have no real roots ... $b^24ac < 0$ $4(k2)^2  8(k2) < 0$ $4(k2)[(k2)  2] < 0$ $4(k2)(k4) < 0 \implies k \in (2,4)$ 
January 12th, 2019, 05:49 AM  #3 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 
Linear Algebra?

January 12th, 2019, 04:12 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,835 Thanks: 733 
For k=2, the "equation" is 2=0, which has no roots. for k=4, the equation becomes $x^2+2x+1=0$ which has a real double root at $x=1$. Last edited by mathman; January 12th, 2019 at 04:16 PM. Reason: add material 
January 13th, 2019, 02:49 PM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 3,044 Thanks: 1627  

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