Question

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

Answer 1

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

Answer 2

x + kx+ 54=0

Answer 3

Cool, now let's do this

Answer 4

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

Answer 5

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

Answer 6

oh sry, x^2

Answer 7

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

Answer 8

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

Answer 9

with me so far?

Answer 10

why are there 2 equations

Answer 11

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

Answer 12

oh ok

Answer 13

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

Answer 14

x^2+kx/2 +27/2

Answer 15

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

Answer 16

why?

Answer 17

Because you can get rid of the annoying x squared now

Answer 18

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

Answer 19

NONONO!

Answer 20

you subtract!

Answer 21

The roots are:\[-3\sqrt{3}, -6\sqrt{3}\] . This means:\[k=9\sqrt{3}\] Now to verify that:
\[(x + 6 \sqrt{3}) (x + 3\sqrt{3}) = x ^{2}+ 9\sqrt{3}+54\]
\[(x +6\sqrt{3})=0\]\[x=-6\sqrt{3}\]\[x = -3\sqrt{3}\]

Answer 22

@radar, show steps don't just give answers

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

=0

Answer 27

then what??

Answer 28

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

Answer 29

oh i forgot it's 27/2

Answer 30

then what

Answer 31

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

Answer 32

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

Answer 33

@phi @Greg_D @mathmate

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

Howw?

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

@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 \(x_- = 2x_+\)

Answer 44

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

Answer 45

@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 \(k=15.6\)

Answer 46

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

Answer 47

lol

Answer 48

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.