I'm currently doing a physics problem and I'm stuck on a bit of algebra. Basically I have the following equation
$$\dfrac{K}{2}\Bigg{[}\dfrac{(4T_{HW}T_H - 4T_{HW}^2 -T_{H}^2 + T_H T_C)}{(2T_{HW} - T_H)^2}\Bigg{]}=0$$
I need to find $T_{HW}$. I tried replacing the variables with letters to make it easier
$$\dfrac{(4ab - 4a^2 -b^2 + bc)}{(2a - b)^2}=0$$
This looks like a quadratic formula problem. I've tried and tried to solve it but I cannot get the result that I need to prove which is $$T_{HW} = \frac{1}{2}(T_H + \sqrt{T_HT_C })$$
or
$$a = \frac{1}{2}(b + \sqrt{bc })$$
Basically I just need a bit of help with the algebra. I know I'm missing something fairly basic but it's really bugging me.
Correcting for amWhy's note, your equation should be $$\frac{4ab - 4a^2 -b^2 + bc}{(2a - b)^2}=0,$$ note that the LHS is undefined for $2a = b$, so we must assume $a \ne b/2$.
With that assumption, you have the fraction turn to zero if and only if the numerator is zero, hence you need $$0 = 4ab - 4a^2 -b^2 + bc = -4a^2+a(4b)+(bc-b^2),$$ which by the quadratic formula yields $$ \begin{split} a_\pm &= \frac{-4b \pm \sqrt{(4b)^2-4\cdot(-4)\cdot(bc-b^2)}}{2\cdot(-4)}\\ &= \frac{-4b \pm \sqrt{(4b)^2+16bc-16b^2}}{-8}\\ &= \frac{b \pm \sqrt{bc}}{2}\\ \end{split} $$ Can you finish simplifying this?