Question 391840
{{{T=-0.005^2 + 0.45x + 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=-0.005^2 + 0.45x + 125}}}
has a solution. Subtracting 135 from each side we get:
{{{0=-0.005^2 + 0.45x - 10}}}
The discriminant will tell us if this has a solution:
{{{(0.45)^2 - 4(-0.005)(-10)}}}
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.)<br>
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=-0.005^2 + 0.45x + 125}}}
Subtracting 134 from each side we get:
{{{0 = -0.005^2 + 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 = (-(0.45) +- sqrt((0.45)^2 - 4(-0.005)(-9)))/2(-0.005)}}}
which simplifies as follows:
{{{x = (-(0.45) +- sqrt(0.2025 - 4(-0.005)(-9)))/2(-0.005)}}}
{{{x = (-(0.45) +- sqrt(0.2025 - 0.180))/2(-0.005)}}}
{{{x = (-(0.45) +- sqrt(0.0225))/2(-0.005)}}}
{{{x = (-0.45 +- sqrt(0.0225))/(-0.01)}}}
{{{x = (-0.45 +- 0.15)/(-0.01)}}}
In long form this is:
{{{x = (-0.45 + 0.15)/(-0.01)}}} or {{{x = (-0.45 - 0.15)/(-0.01)}}}
Simplifying each equation we get:
{{{x = (-0.30)/(-0.01)}}} or {{{x = (-0.60)/(-0.01)}}}
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.