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anonymous
 one year ago
In the equation x + kx+ 54 one root is equal to twice the other root. The value(s) of k
is (are):
a)
5.2
15.6
5.2
15.6
22.0
anonymous
 one year ago
In the equation x + kx+ 54 one root is equal to twice the other root. The value(s) of k is (are): a) 5.2 15.6 5.2 15.6 22.0

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caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0Well, x+kx+54 is polynomial, not equation, so where is the equation?

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0Cool, now let's do this

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0the first thing I would do is solve for x as a function of k

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0Wait a sec! this is first order, meaning it's got only one root. are you missing a square?

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0So we have 2 eqa now.\[x ^{2}+kx+54=0\]

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0and \[4x ^{2}+2kx+54=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why are there 2 equations

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0Becuase it says one rt is twice the other, so if one rt is x, the other has to be 2x, plug 2x in into the original eqa, we can get a new eqa that is equally valid

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0now divide the new eqa by 4 on both side, give me the result

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0fantastic, subtract this eqa above from the original one, what do you get?

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0Because you can get rid of the annoying x squared now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@caozeyuan sorry, i had to do something. so now i set them equal to eachother and solve for x?

radar
 one year ago
Best ResponseYou've already chosen the best response.0The roots are:\[3\sqrt{3}, 6\sqrt{3}\] . This means:\[k=9\sqrt{3}\] Now to verify that: \[(x + 6 \sqrt{3}) (x + 3\sqrt{3}) = x ^{2}+ 9\sqrt{3}+54\] \[(x +6\sqrt{3})=0\]\[x=6\sqrt{3}\]\[x = 3\sqrt{3}\]

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0@radar, show steps don't just give answers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait, shouldn't the second equation be 2x^2+2kx+54?

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0well, no. if x is 2x, then x^2 is (2x)^2 which is 4x^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh right, so i get kx/2 +24 after subtracting

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0I got kx+81=0, WTF is going on?!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i forgot it's 27/2

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0then it sucks, cuz our answer is not in the choices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@phi @Greg_D @mathmate

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0My brain is rusty, I haven't had any math class for half an year

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the answer found by @radar is \(9\sqrt{3}\approx 15.59\) which is close to \(15.6\), the second choice

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0But what's wrong w/ my way？！ IDC anyway cuz I finished highschool algebra a long time ago

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@caozeyuan please give me a minute to read the entire post...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont see any problems starting with \(P(x)=0\) and \(P(2x)=0\), but when you substract you may be missing a solution, because the \(x^2\) term dissapears

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0I see, so I guess I have to solve the eqa with quadratic formula then find k?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me try something in paper, if it works i will post it... just 2 minutes please

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maybe we have to use sum and product rules? C/a and b/a

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@caozeyuan exactly that. if we solve directly we get: \[x_{\pm} = \frac{k\pm \sqrt{k^2216}}{2}\] so we can try either: \(x_+ = 2x_\) or \(x_ = 2x_+\)

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0yep, that's my gut feeling actually, but I though subtraction is much easier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@caozeyuan when dealing with quadratic eqs. (or higher order) you have to be careful when you substract and/or divide, you may loose solutions!! now, if we go with: \(x_+=2x_\) we get: \[\frac{k+\sqrt{k^2216}}{2}=k\sqrt{k^2216}\] so \[k=3\sqrt{k^2216}\] or \[k^2=9(k^2216)\] which solves to: \(k=9\sqrt{3}\approx 15.58\) and \(k=9\sqrt{3}\approx 15.58\) so i guess the answer is the second choice \(k=15.6\)

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0cool, I got that, but I don't care LOL

radar
 one year ago
Best ResponseYou've already chosen the best response.0Sorry about that. I was using the factor method to solve your quadratic equation. The two factors would be of the form (x +a)(x+b) The two factors when multiplied would have to equal your equation\[x ^{2}+kx+54=0\]The product of the numerical terms would have to equal 54. And as stated by the problem b would equal 2b ("one root is equal to twice the other root" and a + 2b would have to equal k.
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