Question

find value of x
Answer a question

Answer 1

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Answer 2

Answer:

(x^4 + 2xi) - (3x² + yi) = (3 - 5i) + (1 + 2yi)  

x^4 + 2xi - 3x² - yi = 3 - 5i + 1 + 2yi  

x^4 + 2xi - 3x² - yi = 4 - 5i + 2yi  

x^4 + 2xi - 3x² - yi - 4 + 5i - 2yi = 0  

x^4 + 2xi - 3x² - 3yi - 4 + 5i = 0  

x^4 - 3x² - 4 + 2xi - 3yi + 5i = 0  

(x^4 - 3x² - 4) + i(2x - 3y + 5) = 0  

Condition for the real part:  

x^4 - 3x² - 4 = 0 → let: X = x² →where X ≥ 0  

X² - 3X - 4 = 0  

X² - (4X - X) - 4 = 0  

X² - 4X + X - 4 = 0  

(X² - 4X) + (X - 4) = 0  

X(X - 4) + (X - 4) = 0  

(X - 4)(X + 1) = 0  

First case: (X - 4) = 0 → X - 4 = 0 → X = 4 → x² = 4 → x = ± 2  

Second case: (X + 1) = 0 → X + 1 = 0 → X = - 1 → no possible  

Condition for the imaginary part:  

2x - 3y + 5 = 0  

3y = 2x - 5  

First possibility: x = 2 → 3y = 4 - 2 → 3y = 2 → y = 2/3  

Second possibility: x = - 2 → 3y = - 4 - 2 → 3y = - 6 → y = - 2  

First solution (2 ; 2/3)  

Second solution (- 2 ; - 2)

Step-by-step explanation: