Question

POLYNOMIALS: SOLVING FOR ROOTS AND UNKOWN COEFFICIENT ....help please! (equation is posted below)

Answer 1

"Two roots of:
\(
\Huge\color{blue} {2x^3 +13x^2+22x+k=0}
\)
are reciprocals of eachother. Find k and all roots of the equation.

Answer 2

latex addict :P

Answer 3

Shhh...

Answer 4

hahaha

Answer 5

say, the roots are : \(\alpha, \dfrac{1}{\alpha}, \beta\)

Answer 6

ok then \[P()=(x-\alpha) (x-\beta)(x \alpha-1)\]

Answer 7

\(\large 2x^3 + 13x^2 + 22x+k
= 2(x-\alpha)(x-\dfrac{1}{\alpha})(x-\beta)\)

Answer 8

Ooohh... ok so I have three unknows therefore i need three equations?

Answer 9

comparing constant terms both sides, you wud get :

\(\large -2\alpha \dfrac{1}{\alpha} \beta = k\)

Answer 10

so -2b=K ...1

Answer 11

that gives you :
\(\large k = -2 \beta\)

Answer 12

yep gotcha :)

Answer 13

Could I do P(b) =0 to find b or ...?

Answer 14

compare x coefficients both sides :
\(\large 22= -2(\alpha + \dfrac{1}{\alpha} + \beta ) \)

compare x^2 coefficients both sides :
\(\large 13 = 2(1 + \dfrac{\beta}{\alpha} + \alpha \beta)\)

Answer 15

you can solve these easily for \(\alpha\) and \(\beta\)

Answer 16

ooohh oK, im trying it now, THANKS @ganeshie8 !!

Answer 17

maybe il get u started a bit, wait it can be tricky if u dont start correctly..

Answer 18

hmm, for your second equation I got \[13=2(1+\frac{ \beta }{ \alpha }- \alpha \beta)\] not PLUS ab...

Answer 19

Hmm, I'm not sure those equations are correct. I've solved the problem and my values of alpha and beta satisfy the constraints but not those equations...

Answer 20

You have below equations to solve :
\(\large 13= -2(\alpha + \dfrac{1}{\alpha} + \beta ) \) ---------(1)

\(\large 22= 2(1 + \dfrac{\beta}{\alpha} + \alpha \beta) \) -------(2)


from the equaiton (1),
\(\large \alpha + \dfrac{1}{\alpha} = -\dfrac{13}{2} - \beta\)


plug this value in equaiton (2) :
\(\large 22= 2(1 + \beta(-\dfrac{13}{2}-\beta))\)
you can sovle this, just a quadratic

Answer 21

ahh flipped x^2 and x coefficients earlier,
corrected now.. thanks @whpalmer4 :) plz check if they're ok now :)

Answer 22

@ganeshie8 I got negatives every time you got a positive for the above equations... @whpalmer4 can you see if your values satisfy the equations when there is a negative""?

Answer 23

note that there are two solutions to that quadratic, and one of them gives you a number that doesn't give you reciprocal roots :-)

Answer 24

which quadratic are you referring to @whpalmer4 ?

Answer 25

\[\large 22= 2(1 + \beta(-\dfrac{13}{2}-\beta))\]

Answer 26

the "wrong" value corresponds to a value of \(k=5\)

Answer 27

the "right" value gives you roots of the big polynomial which are all multiples of each other (and so for that matter is \(k\))

Answer 28

Ok, I'll have a go myself and I'll let you know if I have any troube- thanks for your help @ganeshie8 and @whpalmer4 !!! I really appreciate your time :)

Answer 29

solving the 3 equations you got by comparing coefficients :

http://www.wolframalpha.com/input/?i=13%3D+++-2%28%5Calpha+%2B+%5Cdfrac%7B1%7D%7B%5Calpha%7D+%2B+%5Cbeta+%29%2C+13%3D+++-2%28%5Calpha+%2B+%5Cdfrac%7B1%7D%7B%5Calpha%7D+%2B+%5Cbeta+%29%2C++k+%3D++-2+%5Cbeta

gives k = 8 or 18

Answer 30

Since \(k/2\) is the product of three roots and \(1\) is the product of two of them (they are reciprocals), we can claim that \(k/2\) is a root.

Answer 31

A different approach, if you're going to use Wolfram technology is just to ask to have the various possible polynomials factored :-)

it turns out few negative values of \(k\) give you a factorable polynomial, but there are a number of positive values which will give you a polynomial with rational roots. 0,3,5,8,11,18,35, 96, 143, etc.

Yes, \(k/2\) is indeed a root, and that's a good observation.

Answer 32

actually, it's not a perfect observation, though, because the sign is wrong :-)

Answer 33

hahah im scratching my head around whether complex root pair can be reciprocals of each other

Answer 34

Can anyone explain why \(\dfrac{k}{2}\) is not the product of roots?

Answer 35

when k = 18, this is giving complex solutions... but they're not reciprocals.. hmm

Answer 36

product of roots = -k/2 @idkihavenoidea

Answer 37

well, of course a complex conjugate root pair can be reciprocal — just make \(a =0\) :-)

Answer 38

@ganeshie8 when k is 18 the roots are -4.5, -1+i and -1-i => they are reciprocals!

Answer 39

Oh - for cubics the product is \(-d/a\). Then it's \(-k/2\), yes.

Answer 40

@BarbaraKara sorry, those are not reciprocals...

Answer 41

yeahh lol xD, only when a = 0 i guess...
but why k=18 is not giving reciprrocal roots...

Answer 42

whoops - conjugates i mean..

Answer 43

\[\dfrac{1}{1+i} = \dfrac{1-i}{(1 + i)(1 - i)} = \dfrac{1 - i}{2}\]So \(1+i\) and \(0.5 - 0.5 i\) are reciprocals of each other. Just an example.

Answer 44

yes, but they aren't a conjugate pair...

Answer 45

We may, thus, plug in \(-k/2\) into our equation and try to solve for \(k\).

Answer 46

and any polynomial with only real coefficients will only have complex conjugate roots

Answer 47

Yes, I'm aware of that... just dropping an idea.

Answer 48

okay, thought you were responding to this: "whether complex root pair can be reciprocals of each other"

Answer 49

looks like ive entered the equaitons incorrectly into wolfram, 1 sec -.-

Answer 50

in any case, yes, plugging \(x = -k/2\) into \(2x^3+13x^2+22x + k = 0\) does give you an equation with 3 roots, one of which is the one that solves the problem.

Answer 51

lol

Answer 52

\[-\frac{k^3}{4}+\frac{13 k^2}{4}-10 k=0\]

Answer 53

http://www.wolframalpha.com/input/?i=13%3D+++-2%28%5Calpha+%2B+%5Cdfrac%7B1%7D%7B%5Calpha%7D+%2B+%5Cbeta+%29%2C+22%3D+++2%281%2B+%5Cdfrac%7B%5Cbeta%7D%7B%5Calpha%7D+%2B+%5Calpha+%5Cbeta+%29%2C++k+%3D++-2+%5Cbeta

^gives k = 5 and 8

Answer 54

wow ! yes, pluggin x = -k/2 in the polynomial is the best thing to do here xD

Answer 55

The sum of roots is \(-13/2=-6.5\). If we're getting \(k = 5\) or \(k =8\) then the possible roots are \(-2.5\) and \(-4\).

Answer 56

Ok but as @whpalmer4 says,

"the "wrong" value corresponds to a value of k=5" right?

Answer 57

Try to plug both \(-2.5\) and \(-4\) in.

Answer 58

both k = 5 and k = 8 looks like working

Answer 59

plug k = 5 and see if u get some a reciprocal pair or not..

Answer 60

hmmm

Answer 61

No, roots of \[2x^3+13x^2+22x+5=0\] are \(x = -\frac{5}{2}, x = -2\pm\sqrt{3}\)

Answer 62

oops, sorry, @BarbaraKara don't read that, only @ganeshie8 :-)

Answer 63

ok no worries :)

Answer 64

Ahh, irrational conjugates. I had examples of them.

Answer 65

\(\large -2+\sqrt{3} = \dfrac{1}{-2-\sqrt{3}}\) @whpalmer4

Answer 66

Oooh, thats interesting!

Answer 67

I take back my earlier comment about 5 not working!

Answer 68

thats all good :)

Answer 69

trickery tricky this is lol


k= 5 gives :

\(\large \dfrac{-5}{2}, ~-2 - \sqrt{3}, ~\dfrac{1}{-2-\sqrt{3}}\)

Answer 70

However, \(k=8\) is clearly the superior solution, as both roots and \(k\) are multiples of each other, as previously mentioned :-)

Answer 71

hahah thats called solution biasing :P

Answer 72

Though \(k=5\) is better for the head-scratching metric :-)

Answer 73

True that ! xD

Answer 74

One wonders if the original author of the problem realized that there are two valid solutions...

Answer 75

....solution bias, thats a first! haha

Answer 76

knowing the trickster that is my teacher, yes she would know...

Answer 77

perhaps it would be safer to say "realized that the solution is not unique" :-)

Answer 78

then you should definitely submit the k=5 answer and see if you get called on it!

Answer 79

will do!

Answer 80

you're wlcme \(\huge \color{red}{\ddot\smile}\)

Answer 81

\(
\color{red} {THANKS} \color{orange} {~HEAPS~} \color{green} {EVERYONE:~} \color{blue} {You~all~make~the~world~a~better~place!}
\)

Answer 82

Except for me, I just waste tonnes of oxygen everyday :p

Answer 83

haha! - okay then..

Answer 84

\(
\huge \color{red}{\ddot\smile}~~ nice ~one!
\)
@ganeshie8

Answer 85

...i'll stop now, I promise! haha - farewell, until my next question...

Answer 86

wow ! u became good at copy-pasting others latex code so quick ^_^

Answer 87

well i was taught by the best wasn't I? haha

Answer 88

Do I get to do the next question?

Answer 89

why so keen @idkihavenoidea ?

Answer 90

Oh my xD ! that motivates me to try ur other questions lol :)

Answer 91

I've been studying a similar topic

Answer 92

Ooh awesome! I have lots more! hmmm let me see.....

Answer 93

okay, gtg.. . have a good day !

Answer 94

Don't leave!

Answer 95

Okay, here is one I had trouble with: but first, are u familiar with quotients and remainders @idkihavenoidea ?

Answer 96

it's 3AM here, and I need to be up early, so you'll have to make do without me. I'll take a look in the morning and see if anyone has left me anything tractable to solve :-)