+0

The cubic equation 𝑥^3 − 2𝑥 − 3 = 0 has roots 𝛼, 𝛽 and 𝛾. Find a cubic equation with integer coefficients which have roots;

0
269
5

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

Dec 13, 2018

#1
+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
+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
+3

On the assumption that a) is indeed  $$\frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma}$$ we have: Dec 13, 2018
#4
+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
+1

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

Alan  Dec 13, 2018