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Thread for P5 and P6 questions
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
????????? Help please!
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
????????? Help please!
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
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?????
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?????
Fermat
solve for cos@.
Now substitute for cos@ into z + z^-1 = 2cos @
create a quadratic in z.
solve for z.
Now substitute for cos@ into z + z^-1 = 2cos @
create a quadratic in z.
solve for z.
Makes sense! Cheers guv
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
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