Two correct answers conflicting

[BigMoosie]

I had these questions in my 4 unit maths test yesterday, the red question is the one that I got wrong however I cannot see a problem with my working:

 

a) Find the non real roots of x³ + 1 = 0

 

[math]x = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i[/math]

 

b) If h is one of those roots prove that h² = h - 1

 

[math]LHS = h^2 = \frac{1}{4} \pm \frac{\sqrt{3}}{2}i - \frac{3}{4} = \frac{-1}{2} \pm \frac{\sqrt{3}}{2} i = h - 1 = RHS[/math]

 

c) Hence, find the value of (1-h)^6

 

[math](1-h)^6 = 1 - 6h + 15h^2 - 20h^3 + 15h^4 - 6h^5 + h^6[/math]

 

Considering the roots of unity:

 

[math]= 1 - 6h + 15h^2 - 20 + 15h - 6h^2 + 1[/math]

 

[math]= -18 + 9h + 9h^2[/math]

 

[math]= -27[/math]

 

She got an answer of 1. Why is mine wrong?

[matt grime]

Because it is: i presume you're subbing h^2=h-1 in at some point. Why not do it at the start instead of expanding a sextic unnecessarily? (you're algebra is just wrong)

[cosine]

I had these questions in my 4 unit maths test yesterday' date=' the red question is the one that I got wrong however I cannot see a problem with my working:

 

[b']a) Find the non real roots of x³ + 1 = 0[/b]

 

[math]x = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i[/math]

 

b) If h is one of those roots prove that h² = h - 1

 

[math]LHS = h^2 = \frac{1}{4} \pm \frac{\sqrt{3}}{2}i - \frac{3}{4} = \frac{-1}{2} \pm \frac{\sqrt{3}}{2} i = h - 1 = RHS[/math]

 

c) Hence, find the value of (1-h)^6

 

[math](1-h)^6 = 1 - 6h + 15h^2 - 20h^3 + 15h^4 - 6h^5 + h^6[/math]

 

Considering the roots of unity:

 

[math]= 1 - 6h + 15h^2 - 20 + 15h - 6h^2 + 1[/math]

 

[math]= -18 + 9h + 9h^2[/math]

 

[math]= -27[/math]

 

She got an answer of 1. Why is mine wrong?

Yeah since you've proven B you can make

[math](1-h)^{6}[/math]

into

[math](-1)^{6}(h^{2})^{6} = h^{12} = (h^{3})^{4} = (-1)^{4} = 1[/math]

[BigMoosie]

Because it is: i presume you're subbing h^2=h-1 in at some point. Why not do it at the start instead of expanding a sextic unnecessarily? (you're algebra is just wrong)

 

Where is my algebra wrong? The expansion looks correct, then I subtract 3 from any indicies which are greater than 2 because the roots repeat, then I subtract 9 of each root since the roots sum to zero. That is the technique I have been taught to use with roots of unity.

 

cosine, yeah that was the solution she gave me.

 

I would appreciate to know exactly where the error is because even my Maths teacher couldn't work out why my answer was wrong.

[matt grime]

The roots don't repeat with period 3.

 

They aren't cube roots of unity.

 

They are cube roots of -1.

 

h^4=h^3h=-h.

 

They satisfy x^3+1=0, not x^3-1=0.

[BigMoosie]

The roots don't repeat with period 3.

 

They aren't cube roots of unity.

 

They are cube roots of -1.

 

h^4=h^3h=-h.

 

They satisfy x^3+1=0' date=' not x^3-1=0.[/quote']

 

Agh! Why didn't I see that? Thanks for pinpointing the error Matt.