IIT JEE , AIEEE Forum - Mathematics , Algebra

shinee

x+3i/2+iy = 1-i, then the value of (5x-7y)² =

i got 3 answers for y and x, i wonder y we have to take only the real parts

Decoder

compare both sides...
x=1
3/2 + y = -1...
y = -5/2..

put values..get the ans..

shinee

i got the value of x as 7 and also two imaginary solutions and similarly 4 y , i got the value as 5 and also two other imaginary solutions, if we substitute the real values, we get the answer as 0, but y should we not substitute the imaginary values

is my answer (0) right?

konichiwa2x

How did you get so many values?

shinee

sorry 4 not typing the question properly

the q is x+3i/(2+iy) =1-i
and i solved it like this,
(x+3i)(2-iy)/(4+ysquare) =1-i
2x+3y+6i-ixy=4+ysquare-4i-iysquare
2x+3y=4+ysquare (1)
6-xy= -4-ysquare (2)
x=(4+ysquare-3y)/2 from (1)
substituting in (2)
6 -{(4+ysquare-3y)/2}y =-4-ysquare
ycube-5ysquare+4y-20=0
solving
y=(plus or minus)2i or y =5
and x=(plus or minus)3i or x=7
if we substitute y=5 and x=7, then we r getting the value as 0, but y did we neglect the imaginary answers

konichiwa2x

Consider the equation :

This has 2 solutions because you multiplied by (x-4) . x = 4 is a solution YOU introduced into the system.

SImilarly in this problem, you multiplied by

is a solution you introduced to the system.Hence we need to neglect it.

Now consider the denominator of the expression. What happens when

? Can you see why we neglect this solution also?

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