Question

(2+i)x2-(5-i)x+2(1-i)=0 by quadratic equation

Answer 1

(2 + i)x² - (5 - i)x + 2(1 - i) = 0

from quadratic equation, ax² + bx + c = 0

x = {-b ± √(b² - 4ac)}/2a

so, x = [(5 - i) ± √{(5 - i)² - 4.2(1 - i).(2 + i)}]/2(2 + i)

= [(5 - i) ± √{25 + i² - 10i - 8(2 + i -2i - i²)}]/2(2 + i)

we know, i² = -1

= [(5 - i) ± √{25 -1 - 10i - 8(2 - i + 1)}]/2(2 + i)

= [(5 - i) ± √{24 - 10i - 24 + 8i}]/2(2 + i)

= [(5 - i) ± √{-2i}]/2(2 + i)

here, -2i = (1 - i)²

so, √(-2i) = (1 - i)

so, x = [(5 - i) ± (1 - i)]/2(2 + i)

= (3 - i)/(2 + i), 2/(2 + i)

= (3 - i)(2 - i)/(2 + i)(2 - i) , 2(2 - i)/(2 + i)

= (1 - i) , 2(2 - i)/5

= (1 - i) , 4/5 - 2i/5

hence, there are two values of x

i.e., (1 - i) , 4/5 - 2i/5

Answer 2

Step-by-step explanation:

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