(x+a) (-a+x)(x+2)(x-2+3)=16
Answers
x+a) (-a+x)(x+2)(x-2+3)=16
Step-by-step explanation:
The closer you can get to an algebraic solution is using Lambert W function. It is defined as the inverse function of [math]xe^x[/math], and can help you approximate its complex roots through its properties.
The main idea is to bring that expression to the product log form. Here, let’s try.
[math]x^2 = 16^x[/math]
[math]x^2 = (2^4)^x[/math]
[math]x^2 = 2^{4x}[/math]
[math]x = 2^{2x}[/math]
[math]x = (e^{\ln{(2)}})^{2x}[/math]
[math]x = e^{2\ln{(2)}x}[/math]
[math]e^{2\ln{(2)}x}\frac{1}{x} = 1[/math]
[math]xe^{-2\ln{(2)}x} = 1[/math]
[math]-xe^{-2\ln{(2)}x} = -1[/math]
[math]-2\ln{(2)}xe^{-2\ln{(2)}x} = -2\ln{(2)}[/math]
[math]W(-2\ln{(2)}xe^{-2\ln{2}x}) = W(-2\ln{(2)})[/math]
[math]-2\ln{(2)}x = W(-2\ln{(2)})[/math]
[math]x = -\dfrac{W(-2\ln{(2)})}{2\ln{(2)}}[/math]
One obvious solution is [math]x = -\frac{1}{2}[/math]
But now you have [math]x[/math] in a closed-form expression and can find the complex roots of the equation too.