My Math Forum I've got solved assignment, but don't understand something

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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: 2,982 Thanks: 1575 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: 125 Linear Algebra?
 January 12th, 2019, 04:12 PM #4 Global Moderator   Joined: May 2007 Posts: 6,787 Thanks: 708 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
From: Texas

Posts: 2,982
Thanks: 1575

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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top