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solve :(2 + i)x2 - (5 -i)x + 2(1-i) = 0

(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

sq 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 - 4x +10=0

sovle it furthur by quadratic formula...........

thumps up plzz................

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 thank you sooooo much ... :)

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so thups up na plss....

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(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
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This is really helpful for students

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Please find this answer

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(2+i)x²-(5-i)x+2(1-i)=0
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(2+i)x2-(5-i)x+2(1-i)=0
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a= 1
b= -4
c = 10
x=-b +_ root of b square - 4ac / 2a

 
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x=1-i or x=4/5 - 2/5i
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