(y^2)+y+19

Question

factorize this equation

 

Answer 1

Y^2 + Y + 19 =

   Y ^2 + 2 Y/ 2 +1/4  - 1/4  + 19=

 

   (  Y + 1/2) ^2 + 75/4 =

 

     ( Y + 1/2) ^2 = - 75/4

 

    Y + 1/2 = ± ( ±5i√3 /2) = 

 

   ( Y + 1/2 + 5i√3 / 2) ( Y + 1/2 - 5i√3/ 2)    

Answer 2

If you have types it correctly its factors will be complex conjugates
 
use standard quadratic equation (ax2 + bx +c = 0) solution formula
 
x = (-b + sqrt(b2-4ac))/2a
 
x= ( -1+ sqrt(1-76))/2
x= ( -1 + √-75)2
x= (-1 + 3√5 i)/2        √(-1) = i
 
factors are x + (-1 - 3√5i)/2    and x - (-1 + 3√5i)/2
 
:)

Answer 3

y^2 + y + 19 = 0

 

(a + b)^2 = a^2 + 2ab + b^2

 

We'll force pattern on right and replace with pattern on left:

 

y^2 + 2y(1/2) + (1/2)^2 -  (1/2)^2 + 19 = 0

 

(y+1/2)^2 = 1/4 - 76/4 = -75/4 = -3*25/4

 

|y+1/2| = (5i√(3))/2, where i = √(-1)

 

y+1/2 = ±(5i√(3))/2

 

y = -1/2 ± (5i√(3))/2

 

So y^2 + y + 19 =

(y-(-1/2 + (5i√(3))/2))*(y-(-1/2 - (5i√(3))/2))

 

check:

 

(y-(-1/2 + (5i√(3))/2))*(y-(-1/2 - (5i√(3))/2)) =

 

y^2 - y(-1/2 + (5i√(3))/2 - 1/2 - (5i√(3))/2) + (-1/2 + (5i√(3))/2)*(-1/2 - (5i√(3))/2) =

 

y^2 + y + 1/4 - ((5i√(3))/2)^2 =

 

y^2 + y + 1/4 - 75*(i^2)/4 =

 

y^2 + y + 1/4 + 75/4 =

 

y^2 + y + 76/4 =

 

y^2 + y + 19   √

 

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OTOH, if the second + was supposed to be = [same key], then

y^2 + y = 19

Using the same technique we’d get to:

y^2 + 2y(1/2) + (1/2)^2 - (1/2)^2 = 19

(y + 1/2)^2 = 1/4 + 76/4 = 77/4

|y + 1/2| = (√(77))/2

y + 1/2 = ± (√(77))/2

y = -1/2 ± (√(77))/2