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

##### 1 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) = 24t^3-15t^2+200#

We want to find the zeros of

Given an approximate zero

#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

#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

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

If we try

#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

We can also use Newton's formula to find the two complex zeros, by starting with