SOLUTION: A computer can sort x objects in t seconds, as modeled by the function below: t = 0.003x^2 + 0.001x How many objects are required to keep the computer busy for exactly 7 seconds?

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Question 956644: A computer can sort x objects in t seconds, as modeled by the function below:
t = 0.003x^2 + 0.001x
How many objects are required to keep the computer busy for exactly 7 seconds?
Round to the nearest whole object.

(Please help me, thank you)
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A computer can sort x objects in t seconds, as modeled by the function below:
t= 0.003x^2 + 0.001x
How many objects are required to keep the computer busy for exactly 7 seconds?
Round to the nearest whole object.
====
0.003x^2 + 0.001x = 7
Multiply thru by 1000 to get:
3x^2 + x - 7000 = 0
x = 48.13821
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Ans:: 49
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Cheers,
Stan H.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the formula given is t = .003x^2 + .001x

when t = 7, the formula becomes:

7 = .003x^2 + .001x

subtract 7 from both sides of the equation to get:

.003x^2 + .001x - 7 = 0

this is a quadratic equation in standard form.

a = .003
b = .001
c = -7

use the quadratic formula to solve for x.

that formula is x = %28-b+%2B-+sqrt%28b%5E2-4ac%29%29%2F%282a%29

replace b with .001
replace c with -7
replace a with .003

solve for x to get:

x1 = 48.13821001

x2 = -48.47154335

since x represents the number of objects, then x has to be greater than or equal to 0, so x2 is invalid and the only valid solution is x = 48.13821001.

round to the nearest whole number and the answer is 48.

the computer can sort 48 objects in 7 seconds.

the following graph shows the equation of y = .003x^2 + .001x and shows the equation of y = 7.

the intersection of y = .003x^2 + .001x and y = 7 tells you how many objects the computer can sort in 7 seconds.

when you subtract 7 from both sides of the equation of y = .003x^2 + .001x = 7, you get .003x^2 + .001x - 7 = 0.

the graph of the equation y = .003x^2 + .001x - 7 will show you that y = 7 when x = 48.

the graphs show you both zero crossings.

you ignore the negative value of x because that's invalid.

in other words, when you graph two separate equations of y = .003x^2 + .001x and y = 7, then the value of x when y = 7 will be the intersection of the graph of those two equations, and when you graph y = .003x^2 + .001x - 7, then the value of x when y = 7 will be the intersection of the graph of y = .003x^2 + .001x - 7 with the x-axis.

you'll see that in the following two graphs.

first graph is the two equations of y = .003x^2 + .001x and y = 7.

look for the intersection point of the graph of those two equations.

second graph is the equation of y = .003x^2 + .001x - 7.

look for the intersection of the graph of that equation with the x-axis.

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in the graphs, y is the vertical axis and x is the horizontal axis.

y represents the time in seconds.

x represents the number of objects.

the positive value of x is the only intersection point that's valid.