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A Level maths question help please

Hi everyone, can one of you clever people just check the answer to the last part of the question please.

I got X^3 - 6X - 4 = 0

Many thanks

(edited 11 years ago)

Reply 1

DPLSK

Owhite01

Hi everyone, can one of you clever people just check the answer to the last part of the question please.

Something seems off when I attempted verification using WolframAlpha. I haven't personally verified your solution myself though.

Here's the relevant links.

http://www.wolframalpha.com/input/?i=%28x-sec%5E2%28pi%2F12%29%29%28x-sec%5E2%285pi%2F12%29%29%28x-sec%5E2%283pi%2F4%29%29

http://www.wolframalpha.com/input/?i=x%5E3-6x-4%3D0

I hope this helps.

Darren

Reply 2

Owhite01

Hi Darren
Thanks for the help. I forgot to add last night (before I fell asleep) that the cubic equation is solutions for x=t^2. The way I did the question and I'm not sure it's correct but I couldn't think of another way is to use the fact that sec^2x = 1 + tan^2x and thus to put 1+ t^2 for t into their cubic and come up with an equation in t^6, t^4 etc. I then reduced it to a cubic in x by letting t^2 =x. What do you think?
Thanks Paul

Reply 3

DPLSK

Owhite01

Hi Darren,

Thanks for the help. I forgot to add last night (before I fell asleep) that the cubic equation is solutions for x=t^2. The way I did the question and I'm not sure it's correct but I couldn't think of another way is to use the fact that sec^2x = 1 + tan^2x and thus to put 1+ t^2 for t into their cubic and come up with an equation in t^6, t^4 etc. I then reduced it to a cubic in x by letting t^2=x. What do you think?

Thanks Paul

Hi Paul,

I fail to see why that would work.

My approach would be to rearrange

t^3-3t^2-3t+1=0

to get

t^3-3t^2=3t+1

.

Now square both sides of this. You still get the same roots as before as we haven't really done anything.

Taking everything back over to the LHS gives

(t^3-3t^2)^2-(3t+1)^2=0

.

Expanding this and substituting

t^2=s^2-1

will give you a polynomial of degree 6.

Substituting

x=s^2

will give you what you're after.

Darren

(edited 11 years ago)

Reply 4

Owhite01

Hi Darren
I've actually realised that my first approach was incorrect. What I have now done is factorise their cubic use the relationship sec^2=1+tan^2 along with the t solutions and multiplied out to get a new equation x^3-18x^2+48x-32. What do you think?
Thanx Paul

Reply 5

DPLSK

Owhite01

Hi Darren,

I've actually realised that my first approach was incorrect. What I have now done is factorise their cubic, use the relationship sec^2=1+tan^2 along with the t solutions and multiplied out to get a new equation x^3-18x^2+48x-32. What do you think?

Thanx Paul

Well I'm pretty sure that's the right answer. See above for how I did it.

Darren

(edited 11 years ago)

Reply 6

Owhite01

Hi Darren
Thanks for your solution. I did try a method along your lines but it got too complicated. How would you deal with the term in t^5 when substituting for s^2?
Thanx Paul