word problem--100 deer--please help

[JessicaBurrell]

a herd of 100 deer is indroduced to a small island. the herd at first increases rapidly but eventually the food resources of the island dwindle and population declines. suppose that the number N(t) of deer after t years is given by:
N(t)= -t ^4 + 21t^2 + 100. when does the population of deer become extinct on this island? help!!!!!!!! i do not like word problems at all!!!!!!!!!!!!!!!! thanks

 

[galactus]

When does N(t) equal 0?. Solve for the t values that make N(t)=0.

 

[jwpaine]

N(t)= -t ^4 + 21t^2 + 100

Set the equation equal to zero, and solve for time:

\(\displaystyle \L -t ^{4} + 0t^{3} + 21t^{2} + 0t + 100 = 0\)

Use the rational roots test to find a root, and then divide off to a lower degree polynomial.


Best of luck.

 

[JessicaBurrell]

to jwpaine-thanks

i did the root test, what do i do next? thanks a lot for your help.

 

[jwpaine]

root is going to be one of +/- all factors of C divided by all factors of A

lets pick 5 for our first root.

\(\displaystyle \L -(5)^4 + 21(5)^2+100 = 0\)

\(\displaystyle \L 0 = 0\)

So we now know that 5 is a zero: so (x-5) is a root

now divide the polynomial by (x-5) to arrive at a third degree polynomial; (which will have a leading coefficient c of 20)

find a zero for that, and once again divide off it's root. Keep doing this until you have all 4 solutions.

Note: two solutions are going to be imaginary




Cheers,
John

 

[skeeter]

rational root test ... why?

100 + 21t^2 - t ^4 = 0

(25 - t^2)(4 + t^2)) = 0

t = 5

 

[jwpaine]

Or like skeeter said....use the grouping method to factor into two binomials

\(\displaystyle \L -t^{4}+21t^{2}+100 = 0\)

Two factors of 100 that add up to 21: 25 + -4
Write 21t^2 as the new terms...and group!

\(\displaystyle \L (-t^{4} + 25t^{2}) - (4t^{2} + 100) = 0\)

Remove common factors from each group:

\(\displaystyle \L -t^{2}(t^{2} - 25) -4(t^{2} - 25) = 0\)

You now have the binomial \(\displaystyle \L (t^{2} - 25)\) and combining the factored pieces you have the binomial \(\displaystyle \L (-t^{2} - 4)\)

Thus you are left with \(\displaystyle \L (t^{2} - 25)(-t^{2} - 4) = 0\)


Hope this helps!