Question

Solve the CUBIC equation

Answer 1

Solve the cubic equation if one root is double the other root
\[\huge 24x ^{3} -14x^{2} -63x +45 =0 \]

Answer 2

I have solved this half way through but i am stuck

Answer 3

I applied theory of equations

Answer 4

say the roots are \(\alpha, ~2\alpha,~ \beta\)

Answer 5

Shall i show you what i did , I am stuck in

Answer 6

okay

Answer 7

\[\huge x ^{3}-\frac{ 14 }{ 24 }x ^{2} -\frac{ 63 }{ 24 }x + \frac{ 45 }{ 24 } = 0\]

Answer 8

\[\huge 3\alpha + \beta =\frac{ 14 }{ 23 }\]

Answer 9

\[\huge 3\alpha + \beta =\frac{ 14 }{ 23 }\]\[\huge 2\alpha ^{2} + \alpha \beta + 2\alpha \beta = \frac{ -63 }{ 24 }\]

Answer 10

\[\huge 2\alpha ^{2}\beta = \frac{ -45 }{ 24 }\]

Answer 11

After this what should i do
I tried solving the first equation and get the value of beta then i get a huge quadratic how do i solve that

Answer 12

\[\huge 3\alpha + \beta =\frac{ 14 }{ 24 }\]
\[\huge \beta = \frac{ 7 }{ 12 } - 3 \alpha \]

Answer 13

If i substitute this into the second equation I get
\[\huge 2\alpha ^{2} +3\alpha(\frac{ 7 }{ 12 }- 3\alpha) + \frac{ 63 }{ 24 } =0 \]

Answer 14

How do i solve this equation

Answer 15

looks good so far

Answer 16

multiply thru by 24

Answer 17

\[\huge 48\alpha ^{2} + 72\alpha (\frac{ 7 }{ 12 }- 3a) + 63 = 0 \]

Answer 18

yes, the purpose of multiplying thru b 24 is to get rid off of fractions...
simplify the quadratic and get it into standard good looking form :)

Answer 19

I see so now do i use the quadratic formula

Answer 20

unless it factors...

Answer 21

we have to use it

Answer 22

http://www.wolframalpha.com/input/?i=factor+2%5Calpha+%5E%7B2%7D+%2B3%5Calpha%28%5Cfrac%7B+7+%7D%7B+12+%7D-+3%5Calpha%29+%2B+%5Cfrac%7B+63+%7D%7B+24+%7D+%3D0

Answer 23

yeah use quadratic formula if factoring is not obvious... but looks its going to factor nicely...

Answer 24

1/2 , 3/ 4

Answer 25

Thank you for the help :)

Answer 26

np :) do u have a reason for why u need to ditch one value ?

Answer 27

I didn't get you

Answer 28

after solving the quadratic, you get :

\(\large \alpha = \color{red}{-}\frac{1}{2}, ~\frac{3}{4}\)

Answer 29

right ?

Answer 30

Oh yes uh yeah yes! i wrote 1/2 by mistake

Answer 31

and only one of these two works
you need to pick the right one

Answer 32

yeah putting them in the third equation we have to check that

Answer 33

if it satisfies

Answer 34

let me give u the answer :
\(\alpha = -\frac{1}{2}\) will bot work cuz by descartes rule of signs the given cubic polynomial has EXACTLY one negative root. but taking \(\alpha = -\frac{1}{2}\) gives u two negative roots which is a nonsense... so you need to discard -1/2

Answer 35

not*

Answer 36

ok

Answer 37

the actual roots will be : \(\frac{3}{4}, ~2\times \frac{3}{4}, \beta\)