Question

a) $f(x) = x^2 + 4$
b) $f(x) = -x^2 - 2x$
2 mieiscami zerowymi f

Explanation:

Assuming the problem is to find the zeros (roots) of the given quadratic functions. Let's solve each part:

Part a) \( f(x) = x^2 + 4 \)

Step 1: Set \( f(x) = 0 \)

To find the zeros, we set \( f(x) = 0 \), so we have the equation:
\( x^2 + 4 = 0 \)

Step 2: Solve for \( x \)

Subtract 4 from both sides:
\( x^2 = -4 \)
Taking the square root of both sides, we know that in the real number system, the square root of a negative number is not defined. However, in the complex number system, we can write:
\( x = \pm\sqrt{-4} = \pm 2i \) (where \( i \) is the imaginary unit with \( i^2 = -1 \))

Part b) \( f(x) = -x^2 - 2x \)

Step 1: Set \( f(x) = 0 \)

Set \( f(x) = 0 \), so:
\( -x^2 - 2x = 0 \)

Step 2: Factor out \( -x \)

Factor out \( -x \) from the left - hand side:
\( -x(x + 2)=0 \)

Step 3: Apply the zero - product property

The zero - product property states that if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \).
Case 1: \( -x = 0 \) gives \( x = 0 \)
Case 2: \( x + 2 = 0 \) gives \( x=-2 \)

Final Answers:

a) The zeros of \( f(x)=x^2 + 4 \) in the complex number system are \( x = 2i \) and \( x=-2i \); in the real number system, there are no real zeros.
b) The zeros of \( f(x)=-x^2 - 2x \) are \( x = 0 \) and \( x=-2 \)

Answer:

Step 1: Set \( f(x) = 0 \)

Set \( f(x) = 0 \), so:
\( -x^2 - 2x = 0 \)

Step 2: Factor out \( -x \)

Factor out \( -x \) from the left - hand side:
\( -x(x + 2)=0 \)

Step 3: Apply the zero - product property

The zero - product property states that if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \).
Case 1: \( -x = 0 \) gives \( x = 0 \)
Case 2: \( x + 2 = 0 \) gives \( x=-2 \)

Final Answers:

a) The zeros of \( f(x)=x^2 + 4 \) in the complex number system are \( x = 2i \) and \( x=-2i \); in the real number system, there are no real zeros.
b) The zeros of \( f(x)=-x^2 - 2x \) are \( x = 0 \) and \( x=-2 \)