Question

X^2-X+7 = 0
factor this and find the X's in positive complex root form.

Answer 1

X^(2)-X+7=0

X=(-b+-sqrt(b^(2)-4Xc))/(2X) where X^(2)+bX+c=0

X=(-(-1)+-sqrt((-1)^(2)-4(1)(7)))/(2(1))

X=(1+-sqrt((-1)^(2)-4(1)(7)))/(2(1))

X=(1+-3i sqrt(3))/(2(1))

X=(1+-3isqrt(3))/(2)
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X=(1+3isqrt(3))/(2) , (1-3i~(3))/(2)

Answer 2

thank you, but i have one more question, would you mind solving this too ?
Solve by facotring the equation (x-5)^3 + (2x-1)^3 = 0 where x can be real and/or complex number

Answer 3

(x-5)^(3)+(2x-1)^(3)=0

((x-5)+(2x-1))((x-5)^(2)-(x-5)(2x-1)+(2x-1)^(2))=0

(x-5+2x-1) ((x-5)^(2)-(x-5)(2x-1)+(2x-1)^(2))=0

(3x-6) ((x-5)^(2)-(x-5)(2x-1)+(2x-1)^(2))=0

(3x-6) ((x^(2)-10x+25)-(x-5)(2x-1)+(2x-1)^(2))=0

(3x-6) ((x^(2)-10x+25)-(2x^(2)-11x+5)+(2x-1)^(2))=0

(3x-6) ((x^(2)-10x+25)-2x^(2)+11x-5+(2x-1)^(2))=0

(3x-6) ((x^(2)-10x+25)-2x^(2)+11x-5+(4x^(2)-4x+1))=0

(3x-6) (x^(2)-10x+25-2x^(2)+11x-5+4x^(2)-4x+1)=0

(3x-6)(3x^(2)-3x+21)=0

(3(x-2))(3x^(2)-3x+21)=0

3(x-2)(3(x^(2)-x+7))=0

(9)(x-2)(x^(2)-x+7)=0
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(x-2)=0
(x^(2)-x+7)=0

x-2=0

x=2
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x^(2)-x+7=0

x=(-b+-sqrt(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

x=(-(-1)+-sqrt((-1)^(2)-4(1)(7)))/(2(1))

x=(1+-sqrt((-1)^(2)-4(1)(7)))/(2(1))

x=(1+-3i sqrt(3))/(2(1))

x=(1\3i sqrt(3))/(2)
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x=(1+3i sqrt(3))/(2)

x=(1-3i sqrt(3))/(2)
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x=(1+3i~(3))/(2),(1-3i~(3))/(2)
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x= 2 , (1+3i~(3))/(2) , (1-3i~(3))/(2)

Answer 4

THANK YOU