+0  
 
0
133
5
avatar+777 

The cubic equation 𝑥^3 − 2𝑥 − 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+4396 
+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+99122 
+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+27529 
+2

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

 

 Dec 13, 2018
 #4
avatar+4396 
+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+27529 
+1

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

Alan  Dec 13, 2018

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