Question

(2+i)x^2-(5-i)x+2(1-i)=0 solve this quadratic equation......

with easy method and other any method​

Answer 1

Answer:

1 - i

(4 - 2i)/5

Step-by-step explanation:

(2+i)x²-(5-i)x+2(1-i)=0

D = ( -b ± √b² - 4ac )/2a

b = -(5 - i)

c = 2(1 - i)

a = 2 + i

D = √25 + i² - 10i - 4*2(2 - i² - i)

= √ 24 - 10i - 24 + 8i

= √-2i

= 1 - i    or i - 1

(5 - i  + 1 - i )/2(2 + i)

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

= (3 - i) (2 - i) /(5)

= (5 - 5i)/5

= 1 - i

(5 - i  + i - 1 )/2(2 + i)

= (4)/2(2 + i)

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

=(4 - 2i)/5

Answer 2

Answer:

(2+i)x²-(5-i)x+2(1-i) = 0 ⇒ 2x²+ix²-5x+ix+2-2i =0 ⇒ 2x²-5x+2 = -ix²-ix+2i ⇒ (x-4)(x-1) = -i(x²+x-2) ⇒ (x-4)(x-1) = -i(x+2)(x-1) ⇒ (x-4)/(x+2) = -i Squaring both sides We get :- (x-4)²/(x+2)² = i² ⇒ (x-4)² = -(x+2)² ⇒ x²+16-8x = -x²-4-4x ⇒ 2x²+20-4x = 0 ⇒ x²+10-2x = 0 ⇒ x²-2x+10 = 0 Now solve by yourself using quadratic formula Hope it helps You.☺️ x2 - 4x +10=0