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# math help please??? 10 points?

if the quadratic function is y1= (x-a)^2 + b^2, where a, b ∈ ℝ, how would i show that y1 has zeros a +- ib, where i= √ (-1) ??

and i get that you have to set it equal to 0 to get the zeros, but in the normal equation it is (x-a)^2+b..how would i solve it with b^2??

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- 8 years agoFavorite Answer
In this case b^2 is a constant. So it would be like I was saying 2^2 instead of 4. Just solve normally and leave b^2 as it is. And when you need to input numbers just take the square root of b^2 to find b.

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