Question

The roots of the equation \(x^3-7x^2+Cx -8=0\) form a G.P. Find all roots.

Answer 1

Well how do i start?

Answer 2

what do u mean by ..........form a G.P. ?

Answer 3

the roots form a Geometric Progression

Answer 4

u mean the roots are somthing like \(\alpha\) , \(\alpha \beta\) and \(\alpha\beta^2\) ?

Answer 5

I guess so; i can't remember how to do this.

Answer 6

Hmn so we have common ratio as \(\large{\beta}\) right?

Answer 7

*just for example*

Answer 8

yes i think

Answer 9

can the roots be something like \((\alpha
+\beta)(\beta+\gamma)(\gamma+\alpha)\)?

Answer 10

http://www.wolframalpha.com/input/?i=x3%E2%88%927x2%2BCx%E2%88%928%3D0%20&t=crmtb01

"Seems somewhat 'difficult'"

Answer 11

C is constant right? @Omniscience

Answer 12

well the \(C\) has to be solved afterwards; i assume its a constant

Answer 13

u can do some thing like this

\[(x-\alpha)(x-\alpha \beta)(x-\alpha \beta^2)=x^3-3x^2+Cx-8\]

simplify the Left hand side and then compare the coefficients

Answer 14

ok ... so seems that the given polynomial is cubic ..

Answer 15

u got 3 variables.... a , r , C u got to form three equations
assume roots to be a/r , a , ar

Answer 16

@A.Avinash_Goutham I do think that C is ..constant?

Answer 17

-.- then?

Answer 18

well its not a complex; and im not sure how to find the roots..haven't done these in a long time

Answer 19

http://www.wolframalpha.com/input/?i=a%5E3-bx%5E2%2BCx-d%3D0

You have 3 solutions for x, label these: x1, x2, x3

x2/x1=x3/x2 for it to be geometric progression.

Plug in your values and divide, solve simultaneously.

C cannot be 0, and B cannot be 0.

Answer 20

Once you find x1 and x2. Divide them, then

x1*(x2/x1), x2*(x3/x2)^2

and so fourth. I think this is right, I hope this is right :s

Answer 21

well i dont think that its that easy..

Answer 22

any given information about C (integer or...)

Answer 23

nope; i have to find \(C\) after that

Answer 24

\(\alpha\beta = -2\)

Answer 25

I think I'm wrong. It won't be two equations with two unknown, because the values of x will be different.

Answer 26

we know that that the product of roots for cubic polynomial equation like \(ax^3+bx^2+cx+d=0 \) is -d/a

Answer 27

so according to the roots i assumed for ur equation \(\alpha^3 \beta^3=-8 \) so \(\alpha \beta=-2 \)

Answer 28

sorry -d/a=8 \(\alpha \beta=2 \)

Answer 29

wait how did you get \(2\)?

Answer 30

product of roots=-d/a=8

am i right?

Answer 31

i guess so..

Answer 32

well the roots are \(\alpha\) , \(\alpha \beta\) and \(\alpha \beta^2\) and product of this 3 term is \(\alpha^3 \beta^3\)

Answer 33

so \(\alpha \beta=2\)

Answer 34

\(\alpha^3 \beta^3= (\alpha \beta)^3=8 \) then \(\alpha \beta=2\)

Answer 35

ok thanks

Answer 36

yw :)