I have the following equation:


And I need to find the real solution(s) to said equation, but I don't know how. It's for homework and I don't know how to find a solution to the equation, and Newton-Raphson seems like an unlikely solution if I'm going to solve this equation in a mid-term.

I think there's a way to know if an equation has more than one real root, but I don't remember how to, if anyone wants to explain to me if this is true I will be grateful.

I want to learn an "easy" way of solving these kind of equations without the use of computational aids. I don't want the solution but a way to get to the solution and with a simple calculator punch the numbers at the end and find the answer. Any ideas?

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    $\begingroup$ Have you tried anything at all? Like clearing fractions? Or substituting $1+x=t$? $\endgroup$ – Lubin Feb 19 '13 at 16:47
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    $\begingroup$ Why do you think that a simple calculator punch could be enough? We know that there is no formula for the solutions of the fifth degree equation that only involves functions available in a calculator (roots, products, addition, subtractions, divisions). $\endgroup$ – Jyrki Lahtonen Feb 19 '13 at 16:48
  • $\begingroup$ Do you know that $x$ must be rational? $\endgroup$ – nbubis Feb 19 '13 at 16:56
  • $\begingroup$ My calculator has a button to solve that equation. Maybe you just need a better calculator. $\endgroup$ – MJD Feb 19 '13 at 18:00

This looks like an "internal rate of return" question, where usually you want $x>-1$.

When $x>-1$, the function is strictly decreasing as $x$ increases. Clearly, if $x$ is very large, then the value is negative, and when $x=0$ it is $20000$, so you have one positive real root. You can use numerical methods to find the solution - binary search for example.

I don't know what you'd use on a mid-term - that would depend on what calculation tools you were allowed during the midterm.

One quick way is to write $t=\frac{1}{1+x}$ and note that we are solving $g(t)=0$ for some polynomial with $g(1)=20000$ and $g'(1)=730000$. So an estimate is $t\approx 1-\frac{2}{73}$ or $x\approx 0.028$. This is Newton's method, just on a slightly easier formula when done relative to $t$, and it converges quickly because the root $x$ is close to $0$. This might not work in general.

Wolfram alpha gives the root as $x\approx 0.02932$.

  • $\begingroup$ I have some trouble understanding how you infered that because when $x=0$ there was only one positicve real root, could you please expand a bit on that? $\endgroup$ – alvinbaena Feb 19 '13 at 19:53
  • $\begingroup$ Let $f(x)$ be the function. We know that $f(x)$ is decreasing when $x>-1$, so we know that there can be at most $1$ solution. We know that $f(0)$ is positive, and we know for $x$ large that $f(x)$ is negative. Since $f$ is continuous for $x>-1$, we know that $f(x)=0$ must have a solution. $\endgroup$ – Thomas Andrews Feb 19 '13 at 19:57
  • $\begingroup$ I made a mistake when wrinting the ecuation. I corrected it and tried the method you posted. The result was 0.09756 and wolfram tells me it's 0.13394. This is where the "This might not work in general" comes into play? $\endgroup$ – alvinbaena Feb 19 '13 at 20:33
  • $\begingroup$ Definitely, you'd have to take care with Newton's method in that case. Essentially, I did a single step of Newton's method, it will require more steps or some other technique to get a better answer in general. Again, you haven't stated what tools you are allowed to use in the exam to do this sort of problem, but if you are not permitted a strong calculator at the very least, I don't know how you were supposed to do this problem. $\endgroup$ – Thomas Andrews Feb 19 '13 at 20:38
  • $\begingroup$ When you say, "I want an easy way without computational aids," i think you are out of luck. Solving polynomials is hard, and this sort of problem can be relatively arbitrary, although in most "internal rate of return" problems, I think you are assured only one solution for $x>-1$. $\endgroup$ – Thomas Andrews Feb 19 '13 at 20:39

Letting $1+x=t$, multiplying with $t^5$ and sorting, we find $$t^5=\frac{20t^4+37t^3+54t^2+71t+38}{200}.$$ This suggests an iterative method by letting $t_{n+1}=\sqrt[5]{\frac{20t_n^4+37t_n^3+54t_n^2+71t_n+38}{200}}$. A suitable starting value is $t=1$, leading to a solution $t\approx1.0293$ and $x\approx-0.02849$.

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    $\begingroup$ But at $x=0$ the left side is $20000$ and for large $x$ it is negative. So the root should have $x \gt 0$ $\endgroup$ – Ross Millikan Feb 19 '13 at 17:16
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    $\begingroup$ If $1+x=t$ and $t=1.0293$ then $x=0.0293$ I'd assume. $\endgroup$ – Thomas Andrews Feb 19 '13 at 17:29
  • $\begingroup$ Ah, it looks like, while you solved for $t=x+1$, you then decided that $x=\frac{1}{t}-1$ - that is, you converted back as if you set $t=\frac{1}{1+x}$. $\endgroup$ – Thomas Andrews Feb 19 '13 at 18:02

First make the substitution $t=\frac{1}{1+x}$ so you find a polinomial of degree $5$. Then make the derivative to study the function and see that there is only one solution (for $t$) which is between $0$ and $1$. Then use the bisection method to approximate your solution.

  • $\begingroup$ There is no reason $t$ should be between $0$ and $1$. $\endgroup$ – nbubis Feb 19 '13 at 16:57

Let $t = \dfrac{1}{1+x}$. then you get

$138000t^5 + 71000t^4 + 54000t^3 + 37000t^2 + 20000t - 200000$

Which simplifies to $1000f(t)$ where

$f(t) = 138t^5 + 71t^4 + 54t^3 + 37t^2 + 20t - 200 = 0$

Clearly $f'(t) > 0$ for all $t$, so $f(t)$ is strictly increasing. Hence it has exactly one real root.

We see that $f(0) = -200$ and $f(1) = 120$. So that one real root is between $t=0$ and $t=1$.

There are all sorts of iterative methods that will approximate the value of that root to whatever precision you desire.


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