Question

Find the zeros of (x-2)^2+4
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Answer 1

Answer:

(X-2)^2=x^2+4-4x

x^2-4x+8=0

b^2-4ac=16-24=-8

it is not possible it has no real roots

Answer 2

$(x - {2})^{2} + 4$

$= {x}^{2} + {2}^{2} - 4x + 4$

$= {x}^{2} -4x + 8$

Descriminant,∆=b^2-4ac

$= ( - {4})^{2} - 4(1)(8)$

$= 16 - 32$

$= 16$

So, this polynomial has no real zeros.

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