# Stuck on trivial transfiguration

Oh man, I'm even embarrassed to ask about it but how do I get from $b^2+(\frac{b+c}{2})^2=c^2$ to $c=\frac{5b}{3}$? I'm breaking my head on this one and can't get any further than $c^2=b(3b+2c)$ . . . How should I get rid of this c on the right to get the equation with b on the right only?

-
Are you familiar with the quadratic formula? – Alex Becker Apr 29 '12 at 17:41
take what you got and then treat $c$ as a variable (like $x$) and treat $b$ as a parameter. you get a quadratic equation and there is a formula for the roots (see Wikipidia for example for the formula..) – Belgi Apr 29 '12 at 17:45

First, you’ve a small mistake in your algebra:

$$b^2+\left(\frac{b+c}2\right)^2=b^2+\frac{b^2+2bc+c^2}4=\frac54b^2+\frac12bc+\frac14c^2=c^2\;,$$ so $5b^2+2bc=3c^2$. Now treat this as a quadratic equation $3c^2-2bc-5b^2=0$ in $c$, with $b$ as a constant: $$c=\frac{2b\pm\sqrt{4b^2+60b^2}}6=\frac{b\pm4b}3\;,$$ so $c=\dfrac{5b}3$ or $c=-b$. (Of course you can also factor the quadratic, if you notice the easy factorization.)

-
Thank you very much! :) – Straightfw Apr 30 '12 at 14:56

The given expression is the same as $3c^2 - 2bc-5b^2 = 0$ which upon factoring reduces to $$(c+b)(3c-5b) = 0$$ from which you get what you want.

Added: To factor, write $3c^2-2bc-5b^2$ as \begin{align*} 3c^2 -5bc+3bc-5b^2 = (3c^2 -5bc)+(3bc-5b^2 ) & = c(3c-5b)+b(3c-5b)\\ & = (c+b)(3c-5b) \end{align*}

-
You may want to add an explanation of how to get this factoring, it might seem like magic to the OP – Belgi Apr 29 '12 at 17:46
Haha, that's right - would look like magic without the clarifitcation! :) Thank you both very much. – Straightfw Apr 30 '12 at 14:56