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I need to explain what 2 reasons why
f(x)=x^(2) +4 has no real zeros. im very confused anything helps :)

1 Answer

6 votes

Answer:

Answer in explanation

Explanation:

The graph of f does not cross the x-axis. We know this because for any value x, we always get that x^2+4 is positive. x^2 is positive or zero ( 0 or greater than 0) for any x but adding 4 to it makes It 4 or greater than 4.

The discriminant is negative.

That is, b^2-4ac is negative.

Upon comparing ax^2+bx+c to x^2+4 we see that a=1, b=0, and c=4..

Plugging those into b^2-4ac gives

(0)^2-4(1)(4)

=0-16

=-16.

Negative discriminant implies two nonreal zeros so it will not cross the x-axis.

(Positive discriminant implies 2 real and zero discriminant imples 1 real.)

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