To determine how many objects [tex]\( x \)[/tex] are required to keep the computer busy for exactly 9 seconds, we start by setting up the quadratic equation based on the given model:
[tex]\[
t = 0.007x^2 + 0.003x
\][/tex]
Given [tex]\( t = 9 \)[/tex] seconds, substitute [tex]\( t \)[/tex] into the equation:
[tex]\[
9 = 0.007x^2 + 0.003x
\][/tex]
Rearrange the equation to standard quadratic form:
[tex]\[
0.007x^2 + 0.003x - 9 = 0
\][/tex]
Identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] from the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex]:
[tex]\[
a = 0.007, \quad b = 0.003, \quad c = -9
\][/tex]
Next, we calculate the discriminant [tex]\( \Delta \)[/tex] of the quadratic equation using the formula:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
\Delta = (0.003)^2 - 4(0.007)(-9)
\][/tex]
Calculate the discriminant:
[tex]\[
\Delta = 0.000009 + 0.252 = 0.252009
\][/tex]
Since the discriminant is positive, there are two real solutions to the quadratic equation. Use the quadratic formula to find the roots [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
Calculate the roots:
[tex]\[
x_1 = \frac{-0.003 + \sqrt{0.252009}}{2 \times 0.007}
\][/tex]
[tex]\[
x_2 = \frac{-0.003 - \sqrt{0.252009}}{2 \times 0.007}
\][/tex]
Evaluate the expressions:
[tex]\[
x_1 = \frac{-0.003 + 0.502}{0.014} = \frac{0.499}{0.014} \approx 35.64
\][/tex]
[tex]\[
x_2 = \frac{-0.003 - 0.502}{0.014} = \frac{-0.505}{0.014} \approx -36.07
\][/tex]
Since we need the positive root, we select [tex]\( x_1 \approx 35.64 \)[/tex].
Finally, round [tex]\( x_1 \)[/tex] to the nearest whole object:
[tex]\[
x \approx 36
\][/tex]
Therefore, the computer needs approximately 36 objects to stay busy for exactly 9 seconds.
