anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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caozeyuan
  • caozeyuan
Well, x+kx+54 is polynomial, not equation, so where is the equation?
anonymous
  • anonymous
x + kx+ 54=0
caozeyuan
  • caozeyuan
Cool, now let's do this

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More answers

caozeyuan
  • caozeyuan
the first thing I would do is solve for x as a function of k
caozeyuan
  • caozeyuan
Wait a sec! this is first order, meaning it's got only one root. are you missing a square?
anonymous
  • anonymous
oh sry, x^2
caozeyuan
  • caozeyuan
So we have 2 eqa now.\[x ^{2}+kx+54=0\]
caozeyuan
  • caozeyuan
and \[4x ^{2}+2kx+54=0\]
caozeyuan
  • caozeyuan
with me so far?
anonymous
  • anonymous
why are there 2 equations
caozeyuan
  • caozeyuan
Becuase 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
anonymous
  • anonymous
oh ok
caozeyuan
  • caozeyuan
now divide the new eqa by 4 on both side, give me the result
anonymous
  • anonymous
x^2+kx/2 +27/2
caozeyuan
  • caozeyuan
fantastic, subtract this eqa above from the original one, what do you get?
anonymous
  • anonymous
why?
caozeyuan
  • caozeyuan
Because you can get rid of the annoying x squared now
anonymous
  • anonymous
@caozeyuan sorry, i had to do something. so now i set them equal to eachother and solve for x?
caozeyuan
  • caozeyuan
NONONO!
caozeyuan
  • caozeyuan
you subtract!
radar
  • radar
The 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
  • caozeyuan
@radar, show steps don't just give answers
anonymous
  • anonymous
wait, shouldn't the second equation be 2x^2+2kx+54?
caozeyuan
  • caozeyuan
well, no. if x is 2x, then x^2 is (2x)^2 which is 4x^2
anonymous
  • anonymous
oh right, so i get kx/2 +24 after subtracting
anonymous
  • anonymous
=0
anonymous
  • anonymous
then what??
caozeyuan
  • caozeyuan
I got kx+81=0, WTF is going on?!
anonymous
  • anonymous
oh i forgot it's 27/2
anonymous
  • anonymous
then what
caozeyuan
  • caozeyuan
then it sucks, cuz our answer is not in the choices
anonymous
  • anonymous
..................
anonymous
  • anonymous
@phi @Greg_D @mathmate
caozeyuan
  • caozeyuan
My brain is rusty, I haven't had any math class for half an year
anonymous
  • anonymous
the answer found by @radar is \(9\sqrt{3}\approx 15.59\) which is close to \(15.6\), the second choice
caozeyuan
  • caozeyuan
But what's wrong w/ my way?! IDC anyway cuz I finished highschool algebra a long time ago
anonymous
  • anonymous
Howw?
anonymous
  • anonymous
@caozeyuan please give me a minute to read the entire post...
anonymous
  • anonymous
i 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
  • caozeyuan
I see, so I guess I have to solve the eqa with quadratic formula then find k?
anonymous
  • anonymous
let me try something in paper, if it works i will post it... just 2 minutes please
anonymous
  • anonymous
Maybe we have to use sum and product rules? C/a and -b/a
anonymous
  • anonymous
@caozeyuan exactly that. if we solve directly we get: \[x_{\pm} = \frac{-k\pm \sqrt{k^2-216}}{2}\] so we can try either: \(x_+ = 2x_-\) or \(x_- = 2x_+\)
caozeyuan
  • caozeyuan
yep, that's my gut feeling actually, but I though subtraction is much easier
anonymous
  • anonymous
@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: \[\frac{-k+\sqrt{k^2-216}}{2}=-k-\sqrt{k^2-216}\] so \[k=3\sqrt{k^2-216}\] or \[k^2=9(k^2-216)\] 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
  • caozeyuan
cool, I got that, but I don't care LOL
anonymous
  • anonymous
lol
radar
  • radar
Sorry 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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