SOLUTION: If (x+yi)^2=8-6i, find the values of x and y. I approached this by thinking that x=a and y=b. So I just plugged in the numbers to x and y. => (8-6i)(8-6i) => 64-48i-48i+24i^

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: If (x+yi)^2=8-6i, find the values of x and y. I approached this by thinking that x=a and y=b. So I just plugged in the numbers to x and y. => (8-6i)(8-6i) => 64-48i-48i+24i^      Log On


   

Question 863289: If (x+yi)^2=8-6i, find the values of x and y.
I approached this by thinking that x=a and y=b. So I just plugged in the numbers to x and y.
=> (8-6i)(8-6i)
=> 64-48i-48i+24i^2
=> 64-96i+24i^2
=> 64-96i+24-1
=> 64-96i+23
=> 87-96i
But when I looked up the answer in the book it said x=3 and y=-1.
I'm really hoping you can help tell me how to properly solve this problem.
Thanks and have a good day.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

 (x+yi)^2=8-6i

x^2 + 2xyi - y^2 = 8 -6i

the Idea is (and You will see this again..Apples to Apples

   2xyi = -6i

     y = -3/x

x^2 - 9/x^2 = 8

 x^4 - 8x^2 -9 = 0

(x^2 -9)(x^2 + 1)= 0   Using real roots only

    x^2 = 9

    x = ± 3

   y = -3/x