Determine point of maxima and minima of the function Determine point of maxima and minima of the function 
$f(x)=\frac 1 8 \log x -bx +x^2$, $x> 0$, where $b \geq0$ and is a constant.
I found out $f'(x)$ and equated it to $0$ and got it as
$$8 f'(x)= 1/x - 8b+16x=0.$$ 
Upon solving this I got $16x^2-8bx+1=0$
Solving for $x$ I got $x=(b±(b^2-1)^{1/2})/4$
I don't know how to eliminate the term $b$ and got stuck here.
I also tried $f''$ and got it as 
$8 f''= -1/x^2+16$ and tried using the inequalties for $f''>0$ and $f''<0$  and got $x ∈ (-∞,-1/4)∪(1/4,∞)$ for minima condition and $x ∈ (-1/4,1/4)$ for maxima condition but i was not able to make use of it.

You given that $b>0$.
You got that the function has two extreme values at
$$x_{\pm}=\frac{b\pm \sqrt{b^2-1}}4$$
For $0<b<1$, $x_\pm$ is not real, so restrict the discussion to $b\ge1$.
For $b=1$, $x_\pm=\frac14$ and $f''(\tfrac14)=0$, so the second derivative test is inconclusive (use extremum test to decide that this is a saddle point, since $f'''(\tfrac14)\ne0$).
Now for $b>1$, $$f''(x_+)=16-\frac{1}{\left(\frac{1}{4} \left(b+\sqrt{b^2-1}\right)\right)^2}=\frac{16 \left(\left(b+\sqrt{b^2-1}\right)^2-1\right)}{\left(b+\sqrt{b^2-1}\right)^2}>0,\quad b>1,$$
since $\left(b+\sqrt{b^2-1}\right)^2-1 =2 b^2+2 b\sqrt{b^2-1} -2>2 b \sqrt{ b^2-1}>0$. Thus, $x_+$ is  a minimum for any $b>1$.
Next 
$$f''(x_-)=16-\frac{1}{\left(\frac{1}{4} \left(b-\sqrt{b^2-1}\right)\right)^2}=\frac{16 \left(\left(b-\sqrt{b^2-1}\right)^2-1\right)}{\left(b-\sqrt{b^2-1}\right)^2}$$
So we need to check when $\left(b-\sqrt{b^2-1}\right)^2-1<0$ or equivalently
$$b-\sqrt{b^2-1}<1$$
$$b-1<\sqrt{b^2-1}$$
$$b^2-2b + 1 =(b-1)^2<b^2-1$$
$$2-2b <0$$
Finally, $x_-$ is a maximum for all $b>1$.
