SOLUTION: During a manufacturing process, a metal part in a machine is exposed to varying temperature conditions. The manufacturer of the machine recommends that the temperature of the machi

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Question 391840: During a manufacturing process, a metal part in a machine is exposed to varying temperature conditions. The manufacturer of the machine recommends that the temperature of the machine part remain below 135°F. The temperature T in degrees Fahrenheit x minutes after the machine is put into operation is modeled by T=-0.005^2 + 0.45x + 125.

a. Tell whether the temperature of the part will ever reach or exceed 135°F. Use the discriminant of a quadratic equation to decide.

b. If the machine is in operation for 90 minutes before being turned off, how many times will the temperature of the part be 134°F?
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
T=-0.005%5E2+%2B+0.45x+%2B+125
For part a we will be looking to see if T can ever be 135. In mathematical terms, we are looking to see if the equation
135=-0.005%5E2+%2B+0.45x+%2B+125
has a solution. Subtracting 135 from each side we get:
0=-0.005%5E2+%2B+0.45x+-+10
The discriminant will tell us if this has a solution:
%280.45%29%5E2+-+4%28-0.005%29%28-10%29
0.2025 - 0.200
0.0025
The discriminant is positive. This means that there are two real solutions to the equation. So there will be two times where the temperature will be exactly 135 degrees. (Since the equation for temperature is the equation of a parabola that opens downward, between those two times when the temperature is 135, the temperature will be greater than 135.)

The fact that part a has told us that the temperature will be greater than or equal to 135 tells that the temperature will also be greater than or equal to 134. But We need to know exactly when the temperature is 134 so we know if one or both times is less than 90 minutes. So we need a full solution to:
134=-0.005%5E2+%2B+0.45x+%2B+125
Subtracting 134 from each side we get:
0+=+-0.005%5E2+%2B+0.45x+-+9
Then we use the full quadratic formula. (Note: If 134 was a typo and we are supposed to be using 135, then you can use the previously found discriminant to make the formula simplify faster.)
x+=+%28-%280.45%29+%2B-+sqrt%28%280.45%29%5E2+-+4%28-0.005%29%28-9%29%29%29%2F2%28-0.005%29
which simplifies as follows:
x+=+%28-%280.45%29+%2B-+sqrt%280.2025+-+4%28-0.005%29%28-9%29%29%29%2F2%28-0.005%29
x+=+%28-%280.45%29+%2B-+sqrt%280.2025+-+0.180%29%29%2F2%28-0.005%29
x+=+%28-%280.45%29+%2B-+sqrt%280.0225%29%29%2F2%28-0.005%29
x+=+%28-0.45+%2B-+sqrt%280.0225%29%29%2F%28-0.01%29
x+=+%28-0.45+%2B-+0.15%29%2F%28-0.01%29
In long form this is:
x+=+%28-0.45+%2B+0.15%29%2F%28-0.01%29 or x+=+%28-0.45+-+0.15%29%2F%28-0.01%29
Simplifying each equation we get:
x+=+%28-0.30%29%2F%28-0.01%29 or x+=+%28-0.60%29%2F%28-0.01%29
x = 30 or x = 60
As we can see, both times when the temperature will be exactly 134 are less than 90. So the temperature will be exactly 134 twice within 90 minutes.