# Algebra

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How do I find x for this equation? I think I should get e^(something).

3x^2+4lnx+4=0

• Algebra -

There's no algebraic way to do it. A numeric method or a graphical analysis will be required. See

http://www.wolframalpha.com/input/?i=solve+3x^2%2B4lnx%2B4%3D0

• Algebra -

For a complicated equation such as yours, the most powerful method is probably Newton's Method.

let y = 3x^2+4lnx+4
dy/dx = 6x + 4/x

newx = x - y/(dy/dx)

= x - (3x^2+4lnx+4)/(6x + 4/x)
which simplifies to

newx = (3x^2 - 4lnx)/(6x + 4/x)

now use an iteration process, I started with x = 1

x | newx
1 -- .3
.3 -- .336072..
.336072 -- .337716
.337716 -- .337719143
.337719143 -- .337719143
since the last x used returned the same value of x, we cannot improve on that answer.

Amazing that after only 4 interations, I got
x =.337719143 correct to 9 decimal places

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