Question

{2a+b=c, b^2-a=c} what are the values of a and b?

Answer 1

Let's start by solving for a in both equations:

\[a = \frac{c-b}{2}\]\[a = b^2 - c\]
Since these expressions are equal, we get
\[\frac{c-b}{2} = b^2 - c \Rightarrow c - b = 2b^2 - 2c \Rightarrow 2b^2 + b - 3c = 0\]

Then we use the quadratic formula to solve for b:
\[b = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot -3c}}{2\cdot2} = \frac{-1 \pm \sqrt{1 + 24c}}{4}\]

And because \[a = \frac{c-b}{2}\]we finally get
\[a = \frac{c - \frac{-1\pm\sqrt{1+24c}}{4}}{2} = \frac{1 +4c \mp \sqrt{1+24c}}{8}.\]

Answer 2

Yes, apart from specifying which goes with which....