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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 (k-2)x²+2(k-2)x+2=0 doesn't have real solutions? Solution: D<0 D=(2(k-2))²-4*2(k-2) =4(k²-4k+4)-8(k-2) =4k²-16k+16-8k+16 =4k²-24k+32<0 /:4 =k²-6k+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 ... $b^2-4ac < 0$ $4(k-2)^2 - 8(k-2) < 0$ $4(k-2)[(k-2) - 2] < 0$ $4(k-2)(k-4) < 0 \implies k \in (2,4)$ Thanks from topsquark 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 $x=-1$. Thanks from skeeter, topsquark and qu4lizz 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
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Quote:
 Originally Posted by mathman 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 $x=-1$.
yep ... always look at the original equation. Good catch. Tags assignment, solved, understand Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mayak201 Physics 5 April 8th, 2018 01:33 PM illidan5 Number Theory 5 March 24th, 2016 04:31 AM LMW Algebra 1 November 20th, 2012 02:46 PM ron Calculus 1 October 8th, 2011 09:36 AM

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