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Answer:
this equation has one solution.
A + b = 0
ab = √(5)
since b = -a, subst into other equation
ab = √5
a(-a) = √5
-a² = √5
a² = - √5
a =± (5)^(1/4)·i
so our two zeros are (5)^(1/4) · i and -(5)^(1/4) · i.
Our polynomial has equation
difference of squares: (x + y)(x - y) = x² - y² so
your polynomial is
y = x² + √(5)
y = x^2 + sqrt(5)
ab = √(5)
since b = -a, subst into other equation
ab = √5
a(-a) = √5
-a² = √5
a² = - √5
a =± (5)^(1/4)·i
so our two zeros are (5)^(1/4) · i and -(5)^(1/4) · i.
Our polynomial has equation
difference of squares: (x + y)(x - y) = x² - y² so
your polynomial is
y = x² + √(5)
y = x^2 + sqrt(5)