Which of the following is a solution of x2 − 2x = –8? negative 1 minus i square root of 28 1 plus i square root of 28 negative 1 minus i square root of 7 1 plus i square root of 7

The correct answer choices are:
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 [C]:  " 1 minus i square root of 7" ;  { " x = 1 − i√7 " } ;  AND:
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 [D]:  " 1 plus  i  square root of 7" ;   { " x = 1 + i√7 " } .
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Explanation:
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Given:
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              x² − 2x  = - 8  ;
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  →  Add "8" to EACH SIDE of the equation; to write this equation in "quadratic format" ;  that is:

      " ax² + bx + c = 0" ;  {a ≠ 0} ;
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 →   x² − 2x + 8 = -8 + 8 ;

 →   x² − 2x + 8 = 0 ; 

 →   This is equation is in "quadratic format" ; 

That is:  " ax² + bx + c = 0" ;  {a ≠ 0} ;

in which:  

"a = 1 " {implied coefficient of "1";  
                 since "any value; multiplied by "1", is equal to that same value"} ; 

"b = -2" ; 

"c =   8" ;
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The quadratic equation formula is:__________________________________
x = [-b ± √(b²−4ac] / 2a ;
   = ?  (Let us plug in our known values for "a, b, & c"
x = { [-(-2] ± [√[(-2²) − (4*1*8)] } / {2* 1} ;
x = [2 ± √(-2)] / 2 ;

x = [2 ± √(4 − 32)] / 2 ; 

x = [2 ± √(-28)] / 2 ; 

Note:  Technically, "√-28" does not exist; since there is "no such thing as the square root of a negative number" .
           However, in mathematics, we have designated an "imaginary number", symbolized by the lower case letter, "i" ; to mean:  " √-1 " ; 

So;   " √-28 = √-1 * √28 "  = i * √28 ;  or, " i√28 " . 
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 →  "√28 = √4 * √7 =  2√7 ;

 → " i√28 = i* 2√7 = 2i *√7 = 2i√7 ;
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 →  So;  x = [2 ± √(-28)] / 2 ; 

can be rewritten as:

x = 2 ± 2i√7 / 2 ; 

and this can be simplied; since we are dividing by "2";  all of the THREE "2 digits" cancel out to "1" ;

 →  x = 2 ± 2i√7 / 2  ;

       x = 1 ± 1i √7 / 1 ; 
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 →  Rewrite as:  
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       x =  1 ± i√7 ;

So there are two (2) solutions:
     
       x = 1 + i√7 ; and x = 1 − i√7.
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The correct answer choices are:
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 [C]:  " 1 minus i square root of 7" ;  { " x = 1 − i√7 " } ;  AND:
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 [D]:  " 1 plus  i  square root of 7" ;   { " x = 1 + i√7 " } .
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