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math question 1 week 7

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Type of solution of quadratic depends on the value of Discriminant:  b^2 - 4ac
 
  Roots are evaluated by factoring the equation of 
 
   aX^2 + bx + c
 
  a ( X - b/ 2a ) ^2 - ( b^2 - 4ac)/ 4a^2  
 
   X1 , X = -b/2a ±√(b^2 - 4ac) /2a
 
     if   b^2 - 4ac > 0  , then quadratic has 2 real roos
       
               a.  b^2 -4ac is a perfect square then the roots are rational
 
               b . b^2 - 4ac not a perfect square, then the roots are irrational
    if  b^2 - 4ac< 0 , then quadratic has 2 complex roots.
 
      Quadratic always has 2 roots.
         The irrational and complex roots has to appear as conjugate pair for the coefficients of the quadratic
          to be integers.
 
        i.e. 
 
                X = a ± bi 
                
                 X = a ± √b