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

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- schrodinger

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- caozeyuan

Well, x+kx+54 is polynomial, not equation, so where is the equation?

- anonymous

x + kx+ 54=0

- caozeyuan

Cool, now let's do this

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## More answers

- caozeyuan

the first thing I would do is solve for x as a function of k

- caozeyuan

Wait a sec! this is first order, meaning it's got only one root. are you missing a square?

- anonymous

oh sry, x^2

- caozeyuan

So we have 2 eqa now.\[x ^{2}+kx+54=0\]

- caozeyuan

and \[4x ^{2}+2kx+54=0\]

- caozeyuan

with me so far?

- anonymous

why are there 2 equations

- 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

oh ok

- caozeyuan

now divide the new eqa by 4 on both side, give me the result

- anonymous

x^2+kx/2 +27/2

- caozeyuan

fantastic, subtract this eqa above from the original one, what do you get?

- anonymous

why?

- caozeyuan

Because you can get rid of the annoying x squared now

- anonymous

@caozeyuan sorry, i had to do something. so now i set them equal to eachother and solve for x?

- caozeyuan

NONONO!

- caozeyuan

you subtract!

- 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

@radar, show steps don't just give answers

- anonymous

wait, shouldn't the second equation be 2x^2+2kx+54?

- caozeyuan

well, no. if x is 2x, then x^2 is (2x)^2 which is 4x^2

- anonymous

oh right, so i get kx/2 +24 after subtracting

- anonymous

=0

- anonymous

then what??

- caozeyuan

I got kx+81=0, WTF is going on?!

- anonymous

oh i forgot it's 27/2

- anonymous

then what

- caozeyuan

then it sucks, cuz our answer is not in the choices

- anonymous

..................

- anonymous

@phi @Greg_D @mathmate

- caozeyuan

My brain is rusty, I haven't had any math class for half an year

- anonymous

the answer found by @radar is \(9\sqrt{3}\approx 15.59\) which is close to \(15.6\), the second choice

- caozeyuan

But what's wrong w/ my way？！ IDC anyway cuz I finished highschool algebra a long time ago

- anonymous

Howw?

- anonymous

@caozeyuan please give me a minute to read the entire post...

- 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

I see, so I guess I have to solve the eqa with quadratic formula then find k?

- anonymous

let me try something in paper, if it works i will post it... just 2 minutes please

- anonymous

Maybe we have to use sum and product rules? C/a and -b/a

- 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

yep, that's my gut feeling actually, but I though subtraction is much easier

- 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

cool, I got that, but I don't care LOL

- anonymous

lol

- 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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