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C-Dizzle
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 Posted: Fri Jan 20, 2012 2:15 pm Post subject: Need math help please. Will give rep. |
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Curve, C, has equation y = x^2 + (4k+3)x+7
Line, L, has equation y = x + k
k is constant.
Given that L and C intersect at two distinct points show that 4k^2 + 5k - 6 > 0.
I know I'll have to use b^2 -4ac > 0. Not sure if I need to differentiate the curve and I'm not quite sure if I need to combine the two equations.
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SGL
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SF
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| Posted: Fri Jan 20, 2012 2:26 pm Post subject: |
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the answer is 4.
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C-Dizzle
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| Posted: Fri Jan 20, 2012 2:36 pm Post subject: |
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| SGL wrote: | | http://www.wolframalpha.com/ |
It doesn't do things like that.
| SF wrote: | | the answer is 4. |
The answer is '4k^2 + 5k - 6 > 0'  Not sure how to get there.
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bfsdbsdfbdsfb
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| Posted: Fri Jan 20, 2012 3:44 pm Post subject: |
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| Talix wrote: | please note i'm really benz'd right now
x+k=x^2+x(4k+3)+7
x+k=x^2+4kx+3x+7
k=x^2+4kx+2x+7
k-4kx=x^2+2x+7
k(1-4x)=x^2+2x+7
k=(x^2+2x+7)/(-4x+1)
4k^2+5k-6 > 0
4k^2+8k-3k-6 > 0
4k(k+2)-3(k+2) > 0
(4k-3)(k+2) > 0
k > 3/4
k < -2
that's as far as i can get you |
nowhere is it stated that C = L
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bfsdbsdfbdsfb
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| Posted: Fri Jan 20, 2012 4:03 pm Post subject: |
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| Talix wrote: | | The Grim Heaper wrote: | | Talix wrote: | please note i'm really benz'd right now
x+k=x^2+x(4k+3)+7
x+k=x^2+4kx+3x+7
k=x^2+4kx+2x+7
k-4kx=x^2+2x+7
k(1-4x)=x^2+2x+7
k=(x^2+2x+7)/(-4x+1)
4k^2+5k-6 > 0
4k^2+8k-3k-6 > 0
4k(k+2)-3(k+2) > 0
(4k-3)(k+2) > 0
k > 3/4
k < -2
that's as far as i can get you |
nowhere is it stated that C = L |
| Quote: | | Given that L and C intersect |
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:|
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.....
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| Posted: Fri Jan 20, 2012 4:40 pm Post subject: |
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| Talix wrote: | | The Grim Heaper wrote: | | Talix wrote: | please note i'm really benz'd right now
x+k=x^2+x(4k+3)+7
x+k=x^2+4kx+3x+7
k=x^2+4kx+2x+7
k-4kx=x^2+2x+7
k(1-4x)=x^2+2x+7
k=(x^2+2x+7)/(-4x+1)
4k^2+5k-6 > 0
4k^2+8k-3k-6 > 0
4k(k+2)-3(k+2) > 0
(4k-3)(k+2) > 0
k > 3/4
k < -2
that's as far as i can get you |
nowhere is it stated that C = L |
| Quote: | | Given that L and C intersect |
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How at all does that make L = C. Are you fucking retarded?
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SGL
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| Posted: Fri Jan 20, 2012 5:17 pm Post subject: |
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| Sterling Archer wrote: | | Talix wrote: | | The Grim Heaper wrote: | | Talix wrote: | please note i'm really benz'd right now
x+k=x^2+x(4k+3)+7
x+k=x^2+4kx+3x+7
k=x^2+4kx+2x+7
k-4kx=x^2+2x+7
k(1-4x)=x^2+2x+7
k=(x^2+2x+7)/(-4x+1)
4k^2+5k-6 > 0
4k^2+8k-3k-6 > 0
4k(k+2)-3(k+2) > 0
(4k-3)(k+2) > 0
k > 3/4
k < -2
that's as far as i can get you |
nowhere is it stated that C = L |
| Quote: | | Given that L and C intersect |
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How at all does that make L = C. Are you fucking retarded? |
if they intersect it's because they both give the same value
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I'm SirGodlike
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Cryoma
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| Posted: Fri Jan 20, 2012 5:26 pm Post subject: |
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| SGL wrote: | | Sterling Archer wrote: | | Talix wrote: | | The Grim Heaper wrote: | | Talix wrote: | please note i'm really benz'd right now
x+k=x^2+x(4k+3)+7
x+k=x^2+4kx+3x+7
k=x^2+4kx+2x+7
k-4kx=x^2+2x+7
k(1-4x)=x^2+2x+7
k=(x^2+2x+7)/(-4x+1)
4k^2+5k-6 > 0
4k^2+8k-3k-6 > 0
4k(k+2)-3(k+2) > 0
(4k-3)(k+2) > 0
k > 3/4
k < -2
that's as far as i can get you |
nowhere is it stated that C = L |
| Quote: | | Given that L and C intersect |
|
How at all does that make L = C. Are you fucking retarded? |
if they intersect it's because they both give the same value |
...
If hey gave the same value there would be only one line/curve/watever graphed
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SGL
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| Posted: Fri Jan 20, 2012 5:31 pm Post subject: |
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| Cryoma wrote: | | SGL wrote: | | Sterling Archer wrote: | | Talix wrote: | | The Grim Heaper wrote: | | Talix wrote: | please note i'm really benz'd right now
x+k=x^2+x(4k+3)+7
x+k=x^2+4kx+3x+7
k=x^2+4kx+2x+7
k-4kx=x^2+2x+7
k(1-4x)=x^2+2x+7
k=(x^2+2x+7)/(-4x+1)
4k^2+5k-6 > 0
4k^2+8k-3k-6 > 0
4k(k+2)-3(k+2) > 0
(4k-3)(k+2) > 0
k > 3/4
k < -2
that's as far as i can get you |
nowhere is it stated that C = L |
| Quote: | | Given that L and C intersect |
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How at all does that make L = C. Are you fucking retarded? |
if they intersect it's because they both give the same value |
...
If hey gave the same value there would be only one line/curve/watever graphed |
only when they intersect
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I'm SirGodlike
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InternetIsSeriousBusiness
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| Posted: Fri Jan 20, 2012 5:42 pm Post subject: |
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42
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FLAME FLAME FLAME!!!@@@ |
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Cryoma
Member of the Year
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| Posted: Fri Jan 20, 2012 6:18 pm Post subject: |
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| SGL wrote: | | Cryoma wrote: | | SGL wrote: | | Sterling Archer wrote: | | Talix wrote: | | The Grim Heaper wrote: | | Talix wrote: | please note i'm really benz'd right now
x+k=x^2+x(4k+3)+7
x+k=x^2+4kx+3x+7
k=x^2+4kx+2x+7
k-4kx=x^2+2x+7
k(1-4x)=x^2+2x+7
k=(x^2+2x+7)/(-4x+1)
4k^2+5k-6 > 0
4k^2+8k-3k-6 > 0
4k(k+2)-3(k+2) > 0
(4k-3)(k+2) > 0
k > 3/4
k < -2
that's as far as i can get you |
nowhere is it stated that C = L |
| Quote: | | Given that L and C intersect |
|
How at all does that make L = C. Are you fucking retarded? |
if they intersect it's because they both give the same value |
...
If hey gave the same value there would be only one line/curve/watever graphed |
only when they intersect |
They can't be the same and not intersect wtf they overlap
Jfnrnekdk
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gogodr
I post too much
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| Posted: Fri Jan 20, 2012 10:02 pm Post subject: |
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since I assume you made this:
4k^2 + 5k - 6 > 0
and I don't really know how you came to that conclusion.
only because they intersect you can get this
x^2 + (4k+3)x + 7 = x + k
x^2 + (4k+2)x + 7 = k
x^2 + 4kx + 2x + 7 - k = 0
x^2 + (4x -1)k + 7 = 0
(4x -1)k = -7 - x^2
k = -(7 + x^2)/(4x -1)
2 solutions
x = 0 , k = 7
x = -28 , k = 7
how to?
just make roots out of it and you have a + and a - part where you can evaluate
go to:
http://www.wolframalpha.com/input/?i=k+%3D+-%287+%2B+x^2%29%2F%284x+-1%29
Solutions for the variable x:
x = - sqrt(4 k^2+k-7)-2 k
x = sqrt(4 k^2+k-7)-2 k
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