Can u help me to find a root for C (except c = 0) in below equation.
$$ce^{-c}-{10\over5}(1-e^{-c})^2=0$$ by expanding this I got, $$ce^{-c}-2 + 4 e^{-c}-2e^{-2c}=0$$ now grouping, $$(c+4)e^{-c}-2-2e^{-2c}=0 \rightarrow (1) $$ let $$ e^{-c} = x$$ then $$ c = -logx$$ Substituting these values in (1), $$x^2 + (logx - 4) x +1 = 0$$
now if I apply formula to find the root of quadratic equation, (log x -4) term is coming inside the square root and making it complex to find the root, Am I proceeding it right? Is there any other way to find the root of this equation.
If I plot a graph for this the curve is cutting x axis at 0.49, which is one of the root. How to arrive this mathematically
?
You have to solve it numerically - the analytical solution will involve the Lambert-$\Omega$ (aka product log) function.
See this page for details: http://mathworld.wolfram.com/LambertW-Function.html