0

How would you solve this equation for x. Not sure where to start. Thanks in advance.

$$3.5 = (-7e12*x^4) + (-0.9259*x^3) + (7.222*x^2) + (-22.054*x) + 19.757$$

Paul
  • 9
  • Exponentiation? Isn't this a simple polynomial? – Fimpellizzeri Dec 02 '16 at 23:42
  • @Paul I think this would make a lot more sense to people answering it if you could give some context. How did you get this equation? Also does -7e12 mean $-7\times10^{12}$? – Ian Miller Dec 02 '16 at 23:51
  • Does "$7e12$" represent a pure number or the approximate number "$7.\times 10^{12}$" which represents the interval $[6.5,7.5] \times 10^{12}$? I ask because the potentially low precision of the leading order coefficient essentially discards all the other coefficients. – Eric Towers Dec 03 '16 at 00:20

2 Answers2

2

Unless something is wrong with your typesetting(why the e's in the coefficients?), you are just solving a quartic equation. There is a long and extremely complicated formula for the exact answer, to find the roots of any quartic, you can look online to find it. Alternatively, you could just go on Wolfram Alpha and get numerical approximations(I think it will actually give you exact form too).

Retired account
  • 1,080
1

Numerically.

There is a quartic formula, but its unwieldy to use and not very numerically stable.

An example of a numerical method is Newton's method: if you want to find a solution of $f(x)=0$, start with a guess $x_0$ and apply the iteration

$$ x_n = x_{n-1} - \frac{f(x_{n-1})}{f'(x_{n-1})}$$

until convergence.

If $f(x) = a_N x^N + a_{N-1} x^{N-1} + \ldots + a_1 x + a_0$, $f'(x) = N a_N x^{N-1} + (N-1) a_{N-1} x^{N-2} + \ldots + 2 a_2 x + a_1$.

A lot of mathematical programming languages have functions to get the roots of a polynomial -- for example, in Matlab/Octave, there is one called "roots".

Batman
  • 19,390