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-   -   I've got solved assignment, but don't understand something (http://mymathforum.com/algebra/345613-ive-got-solved-assignment-but-dont-understand-something.html)

qu4lizz January 12th, 2019 04:33 AM

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?

skeeter January 12th, 2019 05:38 AM

I do not agree with the given solution.

discriminant must be strictly negative for the quadratic to have no real roots ...

$b^2-4ac < 0$

$4(k-2)^2 - 8(k-2) < 0$

$4(k-2)[(k-2) - 2] < 0$

$4(k-2)(k-4) < 0 \implies k \in (2,4)$

zylo January 12th, 2019 05:49 AM

Linear Algebra?

mathman January 12th, 2019 04:12 PM

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 $x=-1$.

skeeter January 13th, 2019 02:49 PM

Quote:

Originally Posted by mathman (Post 604384)
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.


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