Maths (O) 2002 PII Question 4 C part (i)

Johnerr

But confused with this question, Maby i'll post it incase u don;t have papers or doing HL

p and k are real numbers such that p(2+i)+8-ki=5k-3-1

(i) find the value of p and the value of k

This what i did

2p+pi+8-ki=5k-3-1 and i went on a bit further, and i realised i did not get the correct answer, then i looked at solutions and found that p(2+i)+8-ki=5k-3-1
the next line is...............
2p+2i+8-ki=5k-3-i


how does p(2+i) = 2p+2i is it not 2p+pi????

can someone please tell me wots going on??

baby*cham*bell

Johnerr wrote:

But confused with this question, Maby i'll post it incase u don;t have papers or doing HL

p and k are real numbers such that p(2+i)+8-ki=5k-3-1

(i) find the value of p and the value of k

This what i did

2p+pi+8-ki=5k-3-1 and i went on a bit further, and i realised i did not get the correct answer, then i looked at solutions and found that p(2+i)+8-ki=5k-3-1
the next line is...............
2p+2i+8-ki=5k-3-i


how does p(2+i) = 2p+2i is it not 2p+pi????

can someone please tell me wots going on??

p(2+i)+8-ki=5k-3-1

p2+pi+8-ki=5k-3-1

p2+8=5k-3-1 pi-ki=0
8+3+1=5k-2p
12=5k-2p







??????????????????????

R.D

Yeah that one got me too! Hope this helps...

http://www.e-xamit.ie/tutorials/leaving/maths/ordinary/2002/1/04/a3.html

Johnerr

my solutions were worng lol, basta*ds, I'm smarter than the book lol

Crumbs

Johnerr wrote:

p and k are real numbers such that p(2+i)+8-ki=5k-3-1

Sure it isn't p(2+i)+8-ki = 5k-3-i ?
(an i at the end instead of a 1).

That should give you the right answer.

rasher_b2

Crumbs wrote:

Sure it isn't p(2+i)+8-ki = 5k-3-i ?
(an i at the end instead of a 1).

That should give you the right answer.

yea, that's what i though too

Johnerr

sorry it is ment to be an i but i got it sorted now,my solutions were worng