How do you solve #6t^4-5t^3+200t+12000 = 175000# ?

1 Answer
Jul 11, 2017

Answer:

The real solutions are approximately:

#t ~~ 13.000377#

#t ~~ -12.6846#

Explanation:

Given:

#p(t) = 6t^4-5t^3+200t+12000 = 175000#

This is a slightly strange question, in that there is one answer very close to a rational number, but it is not exact. Given that (and the fact that the exact algebraic solutions are horribly complicated), it seems to make sense to use a numerical method to find it...

Let:

#f(t) = p(t)-175000 = 6t^4-5t^3+200t-163000#

Then the derivative of #f(t)# is:

#f'(t) = 24t^3-15t^2+200#

We want to find the zeros of #f(t)# using Newton's method:

Given an approximate zero #a_i#, a better approximation is:

#a_(i+1) = a_i - (f(a_i))/(f'(a_i))#

Applying this formula repeatedly we will get better and better approximations.

Where should we start?

Ignoring the terms in #t^3# and #t#, we need:

#6t^4 ~~ 163000#

So:

#t^4 ~~ 163000/6 ~~ 27000#

Then (for real valued solutions at least):

#t^2 ~~ sqrt(27000) ~~ 164#

and

#t ~~ +-sqrt(164) ~~ +-sqrt(169) = +-13#

Trying #a_0 = 13#

We find:

#f(a_0) = 6(color(blue)(13))^4-5(color(blue)(13))^3+200(color(blue)(13))-163000#

#color(white)(f(a_0)) = 171366-10985+2600-163000#

#color(white)(f(a_0)) = -19#

#f'(a_0) = 24(color(blue)(13))^3-15(color(blue)(13))^2+200#

#color(white)(f'(a_0)) = 52728-2535+200#

#color(white)(f'(a_0)) = 50393#

So the next approximation would be:

#a_1 = a_0 - (f(a_0))/(f'(a_0))#

#color(white)(a_1) = 13 - (-19)/50393#

#color(white)(a_1) = 13 + 19/50393#

#color(white)(a_1) ~~ 13.000377#

This approximation is correct to #6# decimal places.

If we try #a_0 = -13# instead we find:

#f(a_0) = 6(color(blue)(-13))^4-5(color(blue)(-13))^3+200(color(blue)(-13))-163000#

#color(white)(f(a_0)) = 171366+10985-2600-163000#

#color(white)(f(a_0)) = 16751#

#f'(a_0) = 24(color(blue)(-13))^3-15(color(blue)(-13))^2+200#

#color(white)(f'(a_0)) = -52728-2535+200#

#color(white)(f'(a_0)) = -55063#

So the next approximation would be:

#a_1 = a_0 - (f(a_0))/(f'(a_0))#

#color(white)(a_1) = -13 - 16751/(-55063)#

#color(white)(a_1) = -13 + 16751/55063#

#color(white)(a_1) ~~ -12.695785#

The actual value is closer to #-12.6846#, so you would need to apply Newton's formula at least one more time for that kind of accuracy.

We can also use Newton's formula to find the two complex zeros, by starting with #a_0 = +-13i#