Halp pls

[Nagisa]

has roots in geometric progression, find a. and hence solve equation.

Now.. i dont know what geometric progression is, well i do but not how to apply it to this.

can anyone help?

[TrueTears]

Assume one root is , thus the other 2 are and where r is some constant ratio.

Now we have

can you go on from here?

[Nagisa]

im trying to get a

[FlorianK]

You basically need to factorise the hole thing into .
The easiest way doing this and solving for is using the horner scheme.
Look up horner scheme and this question will be a piece of cake [EDIT: actually not, but its not too bad]

Try the question I'll post my Tech-active and my tech-free solution in a few hours you should be finished by then or know that you can't solve it for now.

There is a much easier way, but look at the horner scheme anyways

[Nagisa]

You basically need to factorise the hole thing into .
The easiest way doing this and solving for is using the horner scheme.
Look up horner scheme and this question will be a piece of cake.

will do, thanks

[Nagisa]

Assume one root is , thus the other 2 are and where r is some constant ratio.

Now we have

can you go on from here?

why x-x_0,x-rx_0?? (emphasis on the negatives)

[TrueTears]

Spoiler

where coz i felt like using b

Since

so

[Nagisa]

okay true tears, your method works but is long. I realise that what i was doing gives same result

[Planck's constant]

okay true tears, your method works but is long. I realise that what i was doing gives same result

I am all for brief solutions, but you have still not found 'a' nor have you solved the cubic

[Nagisa]

I am all for brief solutions, but you have still not found 'a' nor have you solved the cubic

yeah i know calm down, +1 me and ill do the rest

[Planck's constant]

yeah i know calm down, +1 me and ill do the rest

Deal

[Nagisa]

okay guys, dedicated to argonaut.

okay well, r is obviously 3. so we can plug that back into our a equation. also this means roots = 1, 3, 9

     
     
so
     

[Phy124]

This is how I did it in my head:

Find the factors of 27:

1,3,9,27.

Possible combinations that would multiply to equal 27 could be 1,1,27 or 3,3,3 or 1,3,9 etc.

However we are told the function "has roots in geometric progression", so the only possible combinations are

1,3,9 or 1,-3,9

A little trick to pick up on is that when is expanded it equates to

We know that the coefficient of the term is which is only satisfied when the roots are 1,3,9

Therefore

With all this explaining it might look like rather time consuming but it took no more than 10 seconds to do without a pen or paper

[Nagisa]

This is how I did it in my head:

Find the factors of 27:

1,3,9,27.

Possible combinations that would multiply to equal 27 could be 1,1,27 or 3,3,3 or 1,3,9 etc.

However we are told the function "has roots in geometric progression", so the only possible combinations are

1,3,9 or 1,-3,9

A little trick to pick up on is that when is expanded it equates to

We know that the coefficient of the term is which is only satisfied when the roots are 1,3,9

Therefore

With all this explaining it might look like rather time consuming but it took no more than 10 seconds to do without a pen or paper

except for the finding of the roots part, i did the same thing. I just had to go slow for argonaut's sake, otherwise he wouldn't have +1'd me bro

[curry_bro]

RedHeadedLady, you are a maths goddess what studyscore did u get in methods when you did it? i have a feeling you got 40 scaled at least