The cubic equation 27(z^3) + k(z^2) + 4 = 0 has roots α, β and γ. In the case where β=γ, find the roots of the equation and determine the value of k

This question was worth 8 marks in a 2015 FP2 paper. The first part of this question is tricky, but once we have found the roots it will be easy to plug them in to get the value of k:

When we have an equation of the form 

0 = a(z^3) + b(z^2) + cz + d , with roots α, β and γ, we can write that:

α+β+γ=-b/a     αβ+αγ+βγ=c/a   αβγ=-d/a

The equation in our question has a=27, b=k, c=0 (as there is no z^1 term) and d=4. This gives:

α+β+γ=-k/27     αβ+αγ+βγ=0   αβγ=-4/27

Now we use the information that β=γ, rewriting the second and third equations as:

 2αβ+β^2=0    α(β^2)=-4/27  

subtracting 2αβ from both sides of the left-hand equation gives β^2=-2αβ

Then dividing both sides by β (remembering this means we cannot get β=0) gives β=-2α

Substituting this into the equation α(β^2)=-4/27 gives:

α((-2α)^2)=-4/27

=> α(4(α^2))=-4/27

=> 4α^3=-4/27

=> α^3=-1/27

=> α=-1/3

Almost there. Substituting our value of α back into α(β^2)=-4/27  gives:

-1/3(β^2)=-4/27

=> β^2=4/9

=> β=2/3   Because β=γ, we have also that γ=2/3  

Finally, we bring back our formula for k: α+β+γ=-k/27

substituing in our values for α, β and γ gives (-1/3) + (2/3) + (2/3) = -k/27

=> 1 = -k/27

=> k = -27