Boone Collins
12/20/2023 · Senior High School
Find the zeros of the function \( f(x)=0.3 x^{2}+x-6.6 \). Round values necessary).
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Tutor-Verified Answer
Step-by-step Solution
To find the zeros of the function \( f(x) = 0.3x^2 + x - 6.6 \), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 0.3 \), \( b = 1 \), and \( c = -6.6 \).
First, we calculate the discriminant \( b^2 - 4ac \):
\[
b^2 = 1^2 = 1
\]
\[
4ac = 4 \cdot 0.3 \cdot (-6.6) = -7.92
\]
\[
b^2 - 4ac = 1 - (-7.92) = 1 + 7.92 = 8.92
\]
Now we can substitute \( a \), \( b \), and the discriminant into the quadratic formula:
\[
x = \frac{-1 \pm \sqrt{8.92}}{2 \cdot 0.3}
\]
Calculating \( \sqrt{8.92} \):
\[
\sqrt{8.92} \approx 2.983
\]
Now substituting this value back into the formula:
\[
x = \frac{-1 \pm 2.983}{0.6}
\]
This gives us two solutions:
1. For the positive root:
\[
x_1 = \frac{-1 + 2.983}{0.6} = \frac{1.983}{0.6} \approx 3.305
\]
2. For the negative root:
\[
x_2 = \frac{-1 - 2.983}{0.6} = \frac{-3.983}{0.6} \approx -6.638
\]
Thus, the zeros of the function \( f(x) = 0.3x^2 + x - 6.6 \) are approximately:
\[
x_1 \approx 3.31 \quad \text{and} \quad x_2 \approx -6.64
\]
Rounding to two decimal places, the final answers are:
\[
\boxed{3.31} \quad \text{and} \quad \boxed{-6.64}
\]
Quick Answer
The zeros of the function are approximately \( x_1 = 3.31 \) and \( x_2 = -6.64 \).
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