<s>[INST] what is $a^2+9=b^2+16=1+(a+b)^2$ solve for $a,b$ This is for a geometry question, and through a construction arrived at this equation. I could not solve it and after plugging it into wolfram got the correct answer but can anyone show a method for finding a,b
they are $\frac{5}{\sqrt3}$ and $\frac{2}{\sqrt3}$ respectively (note since this is a geometry question the lengths of a side cannot be negative so the negative solution sets do not matter)
 [/INST] You can start from
$$
1+(a+b)^2=b^2+16 \\
\iff a^2+2ab+b^2+1=b^2+16 \\
\iff a^2+2ab-15=0 \\
\iff b= \frac{15-a^2}{2a}
$$
Now substitute in another equality
$$
a^2+9=b^2+16 \\
\iff a^2+9=\left(\frac{15-a^2}{2a}\right)^2+16 \\
\iff a^2+9=\frac{15^2-30a^2+a^4}{4a^2}+16 \\
\iff 3a^4+2a^2-225 = 0 \\
$$
Solve the quadratic for $a^2$, you get
$$
\implies a^2=\frac{25}{3} \implies a=\frac{5}{\sqrt{3}}
$$
Can you finish from here?
</s>