#1
I have my math final tomorrow and can't seem to contact anyone in my math class at the moment, so I decided to turn to the collective brain of the Pit.

Determine the value(s) of k in the equation so that one root is double the other root:
4x² + kx + 4 = 0

edit: I had written that i need the value of x, but i need the value of k, if that helps
e|----------5-------7-5h7p5---5--
B|-------5----8p5-------------8---8p5--
G|-5h6-----------------------------------7b9r--p5---5b--
D|-----------------------------------------------------7-----7-7~~~
Last edited by AC/DC10 at Jan 28, 2009,
#3
TS, is that one of the opening licks to Texas Flood in your sig?
There may be times when it is impossible to prevent injustice, but there should never be a time when we fail to protest it.


Take a trip down the Scenic River


Call me Charlie.
#5
I failed math. Ha.

The answer is 7.
Ibanez SZ320
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#6
Quote by InvaderTSN
What a coincidence, I have a math final tomorrow too. What are you taking?


Math 11 lol

kinda screwed for the final
e|----------5-------7-5h7p5---5--
B|-------5----8p5-------------8---8p5--
G|-5h6-----------------------------------7b9r--p5---5b--
D|-----------------------------------------------------7-----7-7~~~
#7
at first i thought your sig was the math problem and I was like dude guitar math that rules
Ibanez RG7321
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#8
Quote by AC/DC10
I have my math final tomorrow and can't seem to contact anyone in my math class at the moment, so I decided to turn to the collective brain of the Pit.

Determine the value(s) of k in the equation so that one root is double the other root:
4x² + kx + 4 = 0

edit: I had written that i need the value of x, but i need the value of k, if that helps

your avatar is too distracting
#9
Quote by by_guitar11
I failed math. Ha.

The answer is 7.


actually the answer is +/- 6 square root 2

i just need to find out how to get there
e|----------5-------7-5h7p5---5--
B|-------5----8p5-------------8---8p5--
G|-5h6-----------------------------------7b9r--p5---5b--
D|-----------------------------------------------------7-----7-7~~~
#11
Quote by AC/DC10
actually the answer is +/- 6 square root 2

i just need to find out how to get there



Lol, I threw a random number out there. I'd rather die than attempt to solve that equation.
Ibanez SZ320
Peavey 6505+
Marshall 1960B
Dunlop Wah
Dano EQ
#12
Quote by by_guitar11
Lol, I threw a random number out there. I'd rather die than attempt to solve that equation.

Yeah hahaha
e|----------5-------7-5h7p5---5--
B|-------5----8p5-------------8---8p5--
G|-5h6-----------------------------------7b9r--p5---5b--
D|-----------------------------------------------------7-----7-7~~~
#13
I forget how to do this :P All i nknow is that the discriminant of the equation (b^2-4ac) has to be >0. And a (alpha) = 2b (beta), from the fact that the roots of an equation is a and b (i.e. (x-a)(x-b)=0)

I can't help.
#14
Quote by cHokemeh
I forget how to do this :P All i nknow is that the discriminant of the equation (b^2-4ac) has to be >0. And a (alpha) = 2b (beta), from the fact that the roots of an equation is a and b (i.e. (x-a)(x-b)=0)

I can't help.


Are you saying that i have to use that quadratic formula? or the thing thats like the quadratic formula but not over 2a?
e|----------5-------7-5h7p5---5--
B|-------5----8p5-------------8---8p5--
G|-5h6-----------------------------------7b9r--p5---5b--
D|-----------------------------------------------------7-----7-7~~~
#15
Make sure to use the quadratic equation and keep the k in there, then reduce it and it should make it much easier to see.

I'll at least help you out and give you what it should look like in the quadratic form:

4x² + kx + 4 = 0

x = ( -k ± √[k² - (4)(4)(4)] ) / [(2)(4)]

I don't know what level math you are taking so I don't know how well you can solve that, but if I notice a reply here and you need more help I'll be back to help...IF I'm still on and notice, haha.
I am a lucid dream to the illusionary slumber
Wading in a cesspool of forgotten memory
An insignificant host to the collective subconscious


~Sacred Slumber
#16
try using the quadratic formula then substitute?


or what he said ^^
Quote by Marshmelllow
graphs. graphs always work. my old work place had an awesome printer, so i was constantly making graphs.

that was until i made a graph of how much my boss pissed me off. but seriously dude, graphs.
Last edited by frozen_soul at Jan 28, 2009,
#17
Quote by mhylands
Make sure to use the quadratic equation and keep the k in there, then reduce it and it should make it much easier to see.

I'll at least help you out and give you what it should look like in the quadratic form:

4x² + kx + 4 = 0

x = ( -k ± √[k² - (4)(4)(4)] ) / [(2)(4)]

I don't know what level math you are taking so I don't know how well you can solve that, but if I notice a reply here and you need more help I'll be back to help...IF I'm still on and notice, haha.


YES I think thats just what I needed!
Thanks man!
e|----------5-------7-5h7p5---5--
B|-------5----8p5-------------8---8p5--
G|-5h6-----------------------------------7b9r--p5---5b--
D|-----------------------------------------------------7-----7-7~~~
#18
Haha, no problem. It just might be a little difficult to work out though because of the variable under the square root, but not terribly crazy. Best of luck!
I am a lucid dream to the illusionary slumber
Wading in a cesspool of forgotten memory
An insignificant host to the collective subconscious


~Sacred Slumber
#19
So now I have k±√(k²-24) / 8
What do I do from here?
e|----------5-------7-5h7p5---5--
B|-------5----8p5-------------8---8p5--
G|-5h6-----------------------------------7b9r--p5---5b--
D|-----------------------------------------------------7-----7-7~~~
#20
^(x + 10)(x + 10) FOIL that shit brah.
Quote by AC/DC10
So now I have k±√(k²-24) / 8
What do I do from here?

Would it not be sqrt(k^2) - sqrt(24)? or k - sqrt(24)?

/doesn't know the special codes for maths characters

[IN PHIL WE TRUST]


Quote by Trowzaa
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#21
Quote by SteveHouse
Would it not be sqrt(k^2) - sqrt(24)? or k - sqrt(24)?

/doesn't know the special codes for maths characters


Im not sure, and I just copied and pasted what mhylands said to get the symbols haha
e|----------5-------7-5h7p5---5--
B|-------5----8p5-------------8---8p5--
G|-5h6-----------------------------------7b9r--p5---5b--
D|-----------------------------------------------------7-----7-7~~~
#22
Quote by SteveHouse
FOIL that shit brah.

I think that just made my day.
#23
Quote by AC/DC10
So now I have k±√(k²-24) / 8
What do I do from here?


Well, now you just solve for k.

First of all (4)(4)(4) = 64,

You have k ± √(k²-64) / 8 = x. You'll want to split it up into two equations. You'll have k + √(k²-64) = x and k - √(k²-64) = x. Just from common sense, you know that the latter has to be the smaller of the two, and you know that the two roots you are looking for must be double one another, so you can substitute it as k + √(k²-64) = 2[k - √(k²-64)]. Which further reduces to k + √(k²-64) = 2k - 2√(k²-64). Then you will subtract like terms, so make it so that you have 3√(k²-64) = k. Now you can solve for k by first bring the 3 over, so it'd be √(k²-64) = k/3. Followed by squaring both sides to get things evened up. So it'll be k²-64 = k²/9. Multiply both sides by 9. 9k² - 576 = k². Then move over like terms again, so 8k² = 576. Divide both sides by 8, so k² = 72. Then k = √72 which reduces to k = 6√2. Then you can plug this number into the two separate equations k + √(k²-64) = x and k - √(k²-64) = x to verify the roots are double one another.

As for (x+10)², that is equal to x² + 20x +100.

Hope this helps you out, and that you actually understand it, haha, :P.

EDIT: PS, the k is equal to ±6√2, so you'll have to actually try both +6√2 and -6√2 in the equation to verify which one actually works, but I'm pretty sure it's the positive 6√2.

EDIT2: Haha, didn't realize how many posts got on here before my first post, so yeah I didn't see the post where you put the answer, I guess it is just ±6√2 then XD.
I am a lucid dream to the illusionary slumber
Wading in a cesspool of forgotten memory
An insignificant host to the collective subconscious


~Sacred Slumber
Last edited by mhylands at Jan 28, 2009,
#24
Quote by mhylands
Well, now you just solve for k.

First of all (4)(4)(4) = 64,

You have k ± √(k²-64) / 8 = x. You'll want to split it up into two equations. You'll have k + √(k²-64) = x and k - √(k²-64) = x. Just from common sense, you know that the latter has to be the smaller of the two, and you know that the two roots you are looking for must be double one another, so you can substitute it as k + √(k²-64) = 2[k - √(k²-64)]. Which further reduces to k + √(k²-64) = 2k - 2√(k²-64). Then you will subtract like terms, so make it so that you have 3√(k²-64) = k. Now you can solve for k by first bring the 3 over, so it'd be √(k²-64) = k/3. Followed by squaring both sides to get things evened up. So it'll be k²-64 = k²/9. Multiply both sides by 9. 9k² - 576 = k². Then move over like terms again, so 8k² = 576. Divide both sides by 8, so k² = 72. Then k = √72 which reduces to k = 6√2. Then you can plug this number into the two separate equations k + √(k²-64) = x and k - √(k²-64) = x to verify the roots are double one another.

As for (x+10)², that is equal to x² + 20x +100.

Hope this helps you out, and that you actually understand it, haha, :P.

EDIT: PS, the k is equal to ±6√2, so you'll have to actually try both +6√2 and -6√2 in the equation to verify which one actually works, but I'm pretty sure it's the positive 6√2.


What happens to the denominator?
e|----------5-------7-5h7p5---5--
B|-------5----8p5-------------8---8p5--
G|-5h6-----------------------------------7b9r--p5---5b--
D|-----------------------------------------------------7-----7-7~~~
#25
Quote by AC/DC10
What happens to the denominator?


Haha, sorry, I forgot that. Basically just multiply 8 to the x, so you'll have 8x = EQN(k), and then when I said the whole doubling thing you'd have 8EQN(k₁ = 16EQN(k₂ which reduces to just being EQN(k₁ = 2EQN(k₂ like I had, haha, sorry, hope that clears it up.
I am a lucid dream to the illusionary slumber
Wading in a cesspool of forgotten memory
An insignificant host to the collective subconscious


~Sacred Slumber
Last edited by mhylands at Jan 28, 2009,
#26
Quote by mhylands
You have k ± √(k²-64) / 8 = x. You'll want to split it up into two equations. You'll have k + √(k²-64) = x and k - √(k²-64) = x. Just from common sense, you know that the latter has to be the smaller of the two, and you know that the two roots you are looking for must be double one another, so you can substitute it as k + √(k²-64) = 2[k - √(k²-64)]. Which further reduces to k + √(k²-64) = 2k - 2√(k²-64). Then you will subtract like terms, so make it so that you have 3√(k²-64) = k. Now you can solve for k by first bring the 3 over, so it'd be √(k²-64) = k/3. Followed by squaring both sides to get things evened up. So it'll be k²-64 = k²/9. Multiply both sides by 9. 9k² - 576 = k².


Isnt it -k? So wouldnt that change it somehow? Im getting some funny answers...

Im about to collect like terms, but I have -k+√(k²-64) = -2k - 2√(k²-64)

So then Id get k = -3√(k²-64)

Which goes to -k/3 = √(k²-64)

square everything : k²/9 = k²-64

Multiply by 9: k² = 9k²-576

Like terms: 8k² = 576

Divide by 8 : k² = 72

OMG I GOT IT THANKYOUTHANKYOUTHAKYOU!!!!
e|----------5-------7-5h7p5---5--
B|-------5----8p5-------------8---8p5--
G|-5h6-----------------------------------7b9r--p5---5b--
D|-----------------------------------------------------7-----7-7~~~
Last edited by AC/DC10 at Jan 28, 2009,
#27
Quote by AC/DC10
Isnt it -k? So wouldnt that change it somehow? Im getting some funny answers...


Alright, well I'll actually do it a little more in depth then, haha.

You have k ± √(k²-64) / 8 = x. Multiply both sides by 8 and split the equations. You'll have k + √(k²-64) = 8x and k - √(k²-64) = 8x. Then you can substitute it as 8[k + √(k²-64)] = 2(8)[k - √(k²-64)]. Which you cancel out the 8's and then reduce to k + √(k²-64) = 2k - 2√(k²-64). Then everything goes on the same after that.
I am a lucid dream to the illusionary slumber
Wading in a cesspool of forgotten memory
An insignificant host to the collective subconscious


~Sacred Slumber
Last edited by mhylands at Jan 28, 2009,
#28
Thanks man, I got it. I really appreciate your help and time that you had to put up with me for hahaha. Thanks a lot!
e|----------5-------7-5h7p5---5--
B|-------5----8p5-------------8---8p5--
G|-5h6-----------------------------------7b9r--p5---5b--
D|-----------------------------------------------------7-----7-7~~~
#29
Quote by AC/DC10
Thanks man, I got it. I really appreciate your help and time that you had to put up with me for hahaha. Thanks a lot!


No problem, I'm very glad that I could help out,
I am a lucid dream to the illusionary slumber
Wading in a cesspool of forgotten memory
An insignificant host to the collective subconscious


~Sacred Slumber
#30
Now that this thread is complete... Hey TS, hijack your own thread. Do it. It's bananas.

[IN PHIL WE TRUST]


Quote by Trowzaa
I only play bots. Bots never abandon me. (´・ω・`)