Answer:
The equation D has (1-i) as a solution
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
Option A
in this problem we have
[tex]x^{2} +2x-2=0[/tex]
so
[tex]a=1\\b=2\\c=-2[/tex]
substitute in the formula
[tex]x=\frac{-2(+/-)\sqrt{2^{2}-4(1)(-2)}} {2(1)}[/tex]
[tex]x=\frac{-2(+/-)\sqrt{12}} {2}[/tex]
[tex]x=\frac{-2(+/-)2\sqrt{3}} {2}[/tex]
[tex]x=-1(+/-)\sqrt{3}[/tex]
Has two real solutions
Option B
in this problem we have
[tex]x^{2} +2x+2=0[/tex]
so
[tex]a=1\\b=2\\c=2[/tex]
substitute in the formula
[tex]x=\frac{-2(+/-)\sqrt{2^{2}-4(1)(2)}} {2(1)}[/tex]
[tex]x=\frac{-2(+/-)\sqrt{-4}} {2}[/tex]
Remember that
[tex]i=\sqrt{-1}[/tex]
[tex]x=\frac{-2(+/-)2i} {2}[/tex]
[tex]x=-1(+/-)i[/tex]
[tex]x=-1+i[/tex]
[tex]x=-1-i[/tex]
Option C
in this problem we have
[tex]x^{2} -2x-2=0[/tex]
so
[tex]a=1\\b=-2\\c=-2[/tex]
substitute in the formula
[tex]x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-2)}} {2(1)}[/tex]
[tex]x=\frac{2(+/-)\sqrt{12}} {2}[/tex]
[tex]x=\frac{2(+/-)2\sqrt{3}} {2}[/tex]
[tex]x=1(+/-)\sqrt{3}[/tex]
Has two real solutions
Option D
in this problem we have
[tex]x^{2} -2x+2=0[/tex]
so
[tex]a=1\\b=-2\\c=2[/tex]
substitute in the formula
[tex]x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(2)}} {2(1)}[/tex]
[tex]x=\frac{2(+/-)\sqrt{-4}} {2}[/tex]
Remember that
[tex]i=\sqrt{-1}[/tex]
[tex]x=\frac{2(+/-)2i} {2}[/tex]
[tex]x=1(+/-)i[/tex]
[tex]x=1+i[/tex]
[tex]x=1-i[/tex]
therefore
The equation D has (1-i) as a solution
