+893 Equations like this need a numerical (iterative) method for their solution.
First though is to sketch a graph or graphs in order, if possible, to determine the number and approximate locations of the roots. For this particular example, best is to rewrite the equation as x2 - 4x = 2x and to sketch the graphs of y = x2 - 4x, which is a concave up parabola crossing the x-axis at x=0 and x=4, and y = 2x which looks rather like the y = ex curve. At their point(s) of intersection x2 - 4x = 2x in which case their x co-ordinates will be the solutions to the original equation. There is just one intersection to the left of the origin, located (after trying two or three values for x), round about x = -0.2, and no intersections to the right, (the 2x curve being already at a height of 16 as the parabola becomes positive at x=4).
The most popular numerical method for solving non-linear equations is probably the Newton-Raphson method.
For this equation the formula is
$$x_{n+1}=x_{n}-\frac{x^{2}_{n}-4x_{n}-2^{x_{n}}}{2x_{n}-4-2^{x_{n}}\ln(2)}$$
which, starting with x0 = -0.2 converges to 0.206100 after just two iterations.
It's also usually possible to obtain a convergent iterative formula from the original equation. Rewriting the original equation as x = (x2 - 2x)/4 and using this as an iterative formula, (substitute x=-0.2 on the rhs to get x=-0.2076, substitute this into the rhs etc.), produces the sequence -0.2, -0.2076, -0.2057, -0.2062, -0.20608, etc. Not as fast as Newton-Raphson but it will get there eventually.
+118677 $$x^2-4x-2^x=0$$
$$As\:\:x\rightarrow+\inf \:\:(x^2-4x-2^x)\rightarrow-\inf\\
As\:\:x\rightarrow-\inf \:\:(x^2-4x-2^x)\rightarrow+\inf$$
Therefore there must be an odd number of roots. I'm going to look for just 1
Finding the root of this is not super easy. It can't be factorised.
Let's consider $$f(x)x^2-4x-2^x$$
first I notice that as
$$If \:\: x=0 \mbox{ then } f(x)=0-0-1=-1$$
$$If \:\: x=-1 \mbox{ then } f(x)=1+4-1/2=+4.5$$
So there is at least one root between x=-1 ans x=0
Now use Newton's method of approximating roots to find it.
http://en.wikipedia.org/wiki/Newton's_method
There are you tube videos covering this as well. If you google them.
This is what the graph looks like
+893 Equations like this need a numerical (iterative) method for their solution.
First though is to sketch a graph or graphs in order, if possible, to determine the number and approximate locations of the roots. For this particular example, best is to rewrite the equation as x2 - 4x = 2x and to sketch the graphs of y = x2 - 4x, which is a concave up parabola crossing the x-axis at x=0 and x=4, and y = 2x which looks rather like the y = ex curve. At their point(s) of intersection x2 - 4x = 2x in which case their x co-ordinates will be the solutions to the original equation. There is just one intersection to the left of the origin, located (after trying two or three values for x), round about x = -0.2, and no intersections to the right, (the 2x curve being already at a height of 16 as the parabola becomes positive at x=4).
The most popular numerical method for solving non-linear equations is probably the Newton-Raphson method.
For this equation the formula is
$$x_{n+1}=x_{n}-\frac{x^{2}_{n}-4x_{n}-2^{x_{n}}}{2x_{n}-4-2^{x_{n}}\ln(2)}$$
which, starting with x0 = -0.2 converges to 0.206100 after just two iterations.
It's also usually possible to obtain a convergent iterative formula from the original equation. Rewriting the original equation as x = (x2 - 2x)/4 and using this as an iterative formula, (substitute x=-0.2 on the rhs to get x=-0.2076, substitute this into the rhs etc.), produces the sequence -0.2, -0.2076, -0.2057, -0.2062, -0.20608, etc. Not as fast as Newton-Raphson but it will get there eventually.
