3

I am solving some heat transfer problems, and I came across this equation:

$$(4.536 \cdot 10^{-8})x^4+ 12 x - 4316 = 0$$

The solution is $x = 320$ (I have the solutions book).

I am using a HP50g calculator that gives this same value ($320$), but I need to know how to solve it manually.

Could you help me to understand a step by step of getting to this result? Can Casio scientific calculators solve it?

Andrew Li
  • 4,554
Mila
  • 31
  • There is a complicated formula for finding the roots of fourth degree polynomials - you could look it up. Are you sure you want to do this by hand? Wolfram alpha will do it online for you. – Ethan Bolker Mar 12 '18 at 12:18
  • Actually, I need to know a way to do it without internet or an Hp calculator (using only Casio or manually)... – Mila Mar 12 '18 at 12:22
  • You can use Ferrari's formula for the quartic (https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots). – MrYouMath Mar 12 '18 at 12:24
  • Are you sure the solution is $x=320$? –  Mar 12 '18 at 12:24
  • 2
    Do you allow numerical methods like Newton's? – Parcly Taxel Mar 12 '18 at 12:26
  • @idk yes, I am sure. The solutions manual says it's 320 as well as my HP50g calculator. – Mila Mar 12 '18 at 13:25
  • Well, the solution is not $320$. $(320)^4+12(320)-4316\ne 0$ –  Mar 12 '18 at 13:26
  • @idk Yes, the solution is right, guess you forgot the (4.536⋅10−8) term. ((4.536⋅10−8) x 320^4)+(12 x 320)−4316 = 0 – Mila Mar 12 '18 at 13:29
  • 1
    This quartic equation can't have only one real solution. –  Mar 12 '18 at 13:59
  • @ProfessorVector, you are right, it has 2 real solutions. I am only interested in the positive one, because I am solving a temperature problem in Kelvin (x is the temperature). – Mila Mar 12 '18 at 14:14

2 Answers2

1

You can use Newton's approximation, which I believe is the one used by your calculator too.

Let $f(x)=Ax^4+12x-4316$, where $A=4.536\times10^{-8}$. To approximate $x$, after sufficient iterations, you can use the recursive function below: $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Therefore you get: $$x_{n+1}=x_n-\frac{Ax_n^4+12x_n-4316}{4Ax_n^3+12}$$ After sufficient iterations, you reach $x=320.0203904...$

John Glenn
  • 2,323
  • 11
  • 25
  • 1
    Thanks @John Glenn! That's the algorithm used by the calculator, you are right... – Mila Mar 12 '18 at 14:21
0

Since $A$ is so small, you can use a fixed-point iteration: $$ x_{n+1} = \frac{4316-Ax_n^4}{12} $$ Starting with $x_0=0$ gives $x_{20} \approx 320.0203$.

lhf
  • 216,483