Question

find all zeros for the function: (f(r) = r^2 - 5r + 31). fully simplify.

Explanation:

⚡ Using what you learned: quadratic formula and its applications

Step 1: Set the function to zero

To find the zeros of \( f(r) = r^2 - 5r + 31 \), set \( f(r) = 0 \):

$$ r^2 - 5r + 31 = 0 $$

Step 2: Apply the quadratic formula

Identify the coefficients:

$$ a = 1, \quad b = -5, \quad c = 31 $$

Substitute these values into the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

$$ r = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(31)}}{2(1)} $$

Step 3: Simplify the expression

Simplify the terms under the square root:

$$ r = \frac{5 \pm \sqrt{25 - 124}}{2} $$

$$ r = \frac{5 \pm \sqrt{-99}}{2} $$

Express the square root of the negative number using the imaginary unit \( i \):

$$ \sqrt{-99} = \sqrt{-1 \cdot 9 \cdot 11} = 3i\sqrt{11} $$

Substitute this back into the formula:

$$ r = \frac{5 \pm 3i\sqrt{11}}{2} $$

Answer:

$$ r = \frac{5 \pm 3i\sqrt{11}}{2} $$