Question

question
find the zeros of the function
$f(x) = 1.8x^2 - 16x + 30.5$. round
values to the nearest thousandth (if
necessary).

answer
attempt 1 out of 2

➕ add a zero ➖ no zeros

$x = \square$
submit answer

Explanation:

Step1: Recall Quadratic Formula

For a quadratic function \( f(x) = ax^2 + bx + c \), the zeros are found using \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1.8 \), \( b = -16 \), \( c = 30.5 \).

Step2: Calculate Discriminant (\( D \))

\( D = b^2 - 4ac = (-16)^2 - 4(1.8)(30.5) \)
\( D = 256 - 4(54.9) = 256 - 219.6 = 36.4 \)

Step3: Apply Quadratic Formula

\( x = \frac{-(-16) \pm \sqrt{36.4}}{2(1.8)} = \frac{16 \pm 6.033}{3.6} \)

For \( + \) sign:

\( x_1 = \frac{16 + 6.033}{3.6} = \frac{22.033}{3.6} \approx 6.120 \)

For \( - \) sign:

\( x_2 = \frac{16 - 6.033}{3.6} = \frac{9.967}{3.6} \approx 2.769 \)

Answer:

\( x \approx 2.769 \) and \( x \approx 6.120 \)