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# Thread for P5 and P6 questions watch

1. Post all your P5 and P6 questions here, I know I've got some!!!

For a kick off:
(Solomon, edexcel P6 paper C q2b)

The points A, B and C are (2, 1, -1), (-2, 4, -2) and (a, -5, 1) respectively relative to the origin O, where a isn't 10.
a. Find ABxAC
[I got (10-a)j + 3(10-a)k]

b. The area of triangle ABC is 4.root(10) square units
Find the possible values of a

2. Area of a triangle is (1/2)|ABxAC|
Try using that, and see how you do...
3. See attachment.
Attached Images

4. (Original post by Gaz031)
See attachment.
Cheers gaz, i was being retarded and didn't square the area when i squared the determinant. careless ay, blimey.
thanks again
5. Here's another:

solve 5z^4 - 11z^3 + 16z^2 -11z + 5 = 0
using z^n + z^(-n) = 2cos(nθ)

I've got as far as 5(z^2 + z^-2) - 11(z + z^-1) +16 = 0

so 10cos2θ - 22cosθ + 16 = 0

do you just solve that or what?????
6. solve for [email protected]
Now substitute for [email protected] into z + z^-1 = 2cos @
solve for z.
7. (Original post by Fermat)
solve for [email protected]
Now substitute for [email protected] into z + z^-1 = 2cos @
solve for z.
Makes sense! Cheers guv
8. Solve the quadratic equation in cosθ, noting that cos2θ = 2cos^2θ + 1.
9. (Original post by JamesF)
Solve the quadratic equation in cosθ, noting that cos2θ = 2cos^2θ + 1.
Sorry, I forgot to say it says write the answer in the form a + ib.

so i guess i write z = a + ib and solve the quadratic in that form to get
z = 1/2 + [root(3)/2]i
z = 3/5 + (4/5)i

think that's right, correct me if I'm wrong
10. Sorry, but ...

cos @ = 1/2 or [email protected] = 3/5.

Now substitute for [email protected] into z + 1/z = 2cos @
You should get two values of z ( = a + ib) for each value of [email protected] - four values in all.
It is a quartic eqn after all, so four roots.

Edit:
what you have is correct - just add in the +/- signs.

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