We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
269
5
avatar+844 

The cubic equation π‘₯^3 βˆ’ 2π‘₯ βˆ’ 3 = 0 has roots 𝛼, 𝛽 and 𝛾. Find a cubic equation with integer coefficients which have roots;

For B so far i did:

inverse of 1/x = 1/x 

then i subbed it into the cubic equation however i stopped here as it doesnt match the answer at all 

please help thanks

 Dec 13, 2018
 #1
avatar+6026 
+1

in (b) is it really supposed to be \(\dfrac 1 \alpha + \dfrac 1 \beta\)

 

or does it mean that \(\dfrac 1 \alpha, ~\dfrac 1 \beta,~\dfrac 1 \gamma\) are all roots?

 Dec 13, 2018
 #2
avatar+104394 
+1

I am with Rom, I think that + is meant to be a comma.  

Once you respond to our query we might be able to answer you.

 Dec 13, 2018
 #3
avatar+28159 
+3

On the assumption that a) is indeed  \(\frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma}\) we have:

 

 Dec 13, 2018
 #4
avatar+6026 
+1

or you can just find

 

 \(p\left(\dfrac 1 x\right) = 0\\ \left(\dfrac 1 x\right)^3 - \dfrac 2 x - 3 = 0\\ 1 - 2x^2 - 3x^3 = 0 \\ \text{This is the form the problem asks for, but}\\ \text{dividing both sides by -3}\\ x^3 + \dfrac 2 3 x^2 - \dfrac 1 3 = 0\\ \text{which is the form Alan wrote it in}\)

 

\(\text{similarly for the second part}\\ x=\dfrac{1}{2\alpha+1} \Rightarrow \alpha = \dfrac 1 2\left(\dfrac 1 x -1\right)\\ \text{so find }\\ p\left(\dfrac 1 2\left(\dfrac 1 x -1\right)\right)\\ \text{and work the same sort of algebra to get nice coefficients}\)

Rom  Dec 13, 2018
 #5
avatar+28159 
+1

Yours is a much simpler approach Rom, and, yes, I totally forgot about the "integer coefficients" requirement.

Alan  Dec 13, 2018

5 Online Users

avatar