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find the zeros of the function $f(x) = x^2 - 8.1x + 12.2$. round values…

Question

find the zeros of the function $f(x) = x^2 - 8.1x + 12.2$. round values to the nearest thousandth (if necessary). answer attempt 1 out of 2 add a zero no zeros $x = $

Explanation:

Step1: Recall quadratic formula

For a quadratic function \( f(x) = ax^2 + bx + c \), the zeros are given by \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b=-8.1 \), \( c = 12.2 \).

Step2: Calculate discriminant

Discriminant \( D=b^2 - 4ac=(-8.1)^2-4\times1\times12.2 \)
\( = 65.61-48.8=16.81 \)

Step3: Find roots using quadratic formula

\( x=\frac{-(-8.1)\pm\sqrt{16.81}}{2\times1}=\frac{8.1\pm4.1}{2} \)
For the plus sign: \( x=\frac{8.1 + 4.1}{2}=\frac{12.2}{2}=6.1 \)
For the minus sign: \( x=\frac{8.1-4.1}{2}=\frac{4}{2} = 2 \)

Answer:

The zeros of the function are \( x = 2 \) and \( x=6.1 \) (since the problem has an option to "Add a Zero" and we found two zeros, we can list them. If we need to put one in the box, maybe the first one or either, but the two zeros are \( 2 \) and \( 6.1 \))