Math Question SOLVED
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13 Dec 2009, 12:13
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Journals
Aight, have exams of maths tommorow and I have forgotten how to solve this question. Didn't wrote the solution down :/
x² - (k - 5) - (k + 2) . (2k - 3) = 0
Only thing you can do is to solve it with the discriminant but if I do that, I have sth like this : D = -7k² -14k + 49
Another second degree equation :/ Fucked up.
Tnx for help,
- Piegie
EDIT : Found the solution, but I still don't have any idea how they got there..
x1 = - k + 5 x2 = 2k - 3
x² - (k - 5) - (k + 2) . (2k - 3) = 0
Only thing you can do is to solve it with the discriminant but if I do that, I have sth like this : D = -7k² -14k + 49
Another second degree equation :/ Fucked up.
Tnx for help,
- Piegie
EDIT : Found the solution, but I still don't have any idea how they got there..
x1 = - k + 5 x2 = 2k - 3
x=wurzel aus:2k+2k²-11
Ik denk dat je gwn de k & x² naar de overkant moet doen, en hem van toestandsteken veranderen. en alles normale cijfers aan de linkerkant.
Als het om vergelijkingen gaat.
ik ga zelf niet beginnen denken ;)
<=>x²=k-5+3k²-3k+4k-6
<=>x²-2k²+2k+11=0
x² is always positive so you don't care "x" can be a number from R
Then you have to do the discriminant with -2k²+2k+11=0
I'm not really sure.
you've got this:
x² - (k - 5) - (k + 2) * (2k - 3) = 0
or this:
x² - (k - 5)x - (k + 2) * (2k - 3) = 0
x²= (k - 5) + (k + 2) . (2k - 3)
k = 11
np4me
D = ( k -5 )² - 4. (( k + 2 ) . ( 2k - 3 ))
= k² - 10k + 25 - 4 . ( 2k² + k - 6 )
= k² - 10k + 25 - 8k² - 4 + 24
= -7k² - 14k + 49
En nu graak k nie mee verder :/ Oftewel moest ik het niet distributief uitgewerkt hebben.
'k kan nu ook geen merkwaardig product daarop toepassen. Als het zo was geweest wel : k² - 14k + 49.
Me haz no id.
-(k-5)-((k+2)*(2k-3))=-x²
-k+5-((k+2)*(2k-3))=-x²
-k+5-(2k²-3k+4k-6)=-x²
-k+5-2k²+3k-4k+6=-x²
-2k²+3k-4k-k+5+6=-x²
-2k²-2k+11=-x² |*(-1)
2k²+2k-11=x²
?
x= wurzel aus:2k²+2k-11
you failed at: -k+3k-4k=-2k
Btw, found the answers but still not how to them.
edit:
how to:
x²-k+5-(2k²-3k+4k-6)=0
x²-k+5-2k²+3k-4k+6=0
x²-2k-2k²+11=0 /+2k+2k²-11
x²=2k+2k²-11
-7k(k+2) + 49
(k+2) (49-7k)
k+2=0 v 49-7k=0
k=-2 v -7k=-49 /:(-7)
___________ k=2
? :D
edit: no lol delta doesnt work here.
x² - (k - 5) - (k + 2) * (2k - 3) = 0 //// -x²
- (k - 5) - (k + 2) * (2k - 3) = -x² //// * (-1)
(k - 5) + (k + 2) * (2k - 3) = x² ----> you dont need to do anything just put this (k - 5) + (k + 2) * (2k - 3) in to the main equation, don't make your work more hard just make it as easy as you can.
it will looks like
(k - 5) + (k + 2) * (2k - 3) - (k - 5) - (k + 2) * (2k - 3) = 0 ---> lets make it easier
k - 5 + 2k² -3k +4k -6 - k - 5 - 2k² - 3k + 4k -6 = 0
2k - 5 - 6 -5 -6 = 0
2k - 22 = 0
2k = 22
k = 11
so final check
(11 - 5) + (11 +2) * (11*2 -3) - (11 -5) - (11 + 2) * (11*2 -3) = 0
6 + 247 - 6 - 247 = 0
0 = 0
which means i was right
- (k - 5) = - k + 5 ?
would make k = 6
1 + 72 - 18 + 24 - 6 - 6 + 5 - 72 - 18 + 24 - 6 = 0
- 35 + 48 - 13 = 0
- 35 + 35 = 0
0 = 0