The fish population in a certain part of the ocean (in thousands of fish) as a function of the water's temperature (in degrees Celsius) is modeled by: $p(x)=-2x^2+40x-72$ Which temperatures will result in no fish (i.e. population)?

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Q. The fish population in a certain part of the ocean (in thousands of fish) as a function of the water's temperature (in degrees Celsius) is modeled by:

p(x)=-2x^2+40x-72

Which temperatures will result in no fish (i.e. population)?

Set up quadratic equation: To find the temperatures that result in no fish population, we need to solve the quadratic equation $p(x) = -2x^2 + 40x - 72 = 0$ for $x$ .
Calculate discriminant: The quadratic equation is in the form $ax^2 + bx + c = 0$ , where $a = -2$ , $b = 40$ , and $c = -72$ . We can solve for $x$ by using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$
Apply quadratic formula: First, we calculate the discriminant, which is $b^2 - 4ac$ . For our equation, the discriminant is $(40)^2 - 4(-2)(-72)$ .
Simplify formula: Calculating the discriminant gives us $1600 - 4(-2)(-72) = 1600 - 576 = 1024$ .
Calculate first solution: Since the discriminant is positive, we have two real solutions. Now we can apply the quadratic formula: $x = \frac{-40 \pm \sqrt{1024}}{2 \times -2}$ .
Calculate second solution: Simplifying the quadratic formula gives us $x = \frac{{-40 \pm 32}}{{-4}}$ .
Identify zero fish temperatures: We have two solutions for $x$ : $x = (-40 + 32) / -4$ and $x = (-40 - 32) / -4$ .
Identify zero fish temperatures: We have two solutions for $x$ : $x = (-40 + 32) / -4$ and $x = (-40 - 32) / -4$ .Calculating the first solution: $x = (-40 + 32) / -4 = -8 / -4 = 2$ .
Identify zero fish temperatures: We have two solutions for $x$ : $x = (-40 + 32) / -4$ and $x = (-40 - 32) / -4$ . Calculating the first solution: $x = (-40 + 32) / -4 = -8 / -4 = 2$ . Calculating the second solution: $x = (-40 - 32) / -4 = -72 / -4 = 18$ .
Identify zero fish temperatures: We have two solutions for $x$ : $x = (-40 + 32) / -4$ and $x = (-40 - 32) / -4$ .Calculating the first solution: $x = (-40 + 32) / -4 = -8 / -4 = 2$ .Calculating the second solution: $x = (-40 - 32) / -4 = -72 / -4 = 18$ .The temperatures that result in a fish population of zero are $2$ degrees Celsius and $18$ degrees Celsius.