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)
 A: We have $b^2+2ab-8=0$. Solving this for $a$ and substituting we obtain the following solutions
$$
(a, b)=(3i,-4i),(-3i,4i),(5/\sqrt{3},2/\sqrt{3}),(-5/\sqrt{3},-2/\sqrt{3}).
$$
So the only solution in positive real numbers is $(a, b)=(5/\sqrt{3})$.
A: Maybe try this way:
$b^2+16=1+(a+b)^2\Rightarrow 15=a(a+2b)$ and $a^2+9=1+(a+b)^2\Rightarrow 8=b(2a+b)$
$\frac{8}{15}=\frac{b(2a+b)}{a(a+2b)}=\frac{x(2+x)}{1+2x}$, where $x=\frac{b}{a}$ (Assuming $a$ and $b$ are positive).
On solving, we get $x=\frac{2}{5}=\frac{b}{a}$.
Thus, $5b=2a$. 
As $a^2+9=b^2+16=(\frac{2a}{5})^2+16$. This gives $a=\frac{5}{\sqrt{3}}$. Thus, $b=\frac{2}{\sqrt{3}}$. (Considering only positive numbers)
A: 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?
A: Let
\begin{align} 
a^2+9&=k
\tag{1}\label{1}
,\\
b^2+16&=k
\tag{2}\label{2}
,\\
1+(a+b)^2&=k
\tag{3}\label{3}
\end{align}
for some $k>0$.
Then \eqref{3}$-$\eqref{1}$-$\eqref{2} gives
\begin{align}
ab &= 12-\tfrac12\,k
\tag{4}\label{4}
,\\
a^2b^2 &= (12-\tfrac12\,k)^2
\tag{5}\label{5}
.
\end{align}
On the other hand, from \eqref{1},\eqref{2}
we have
\begin{align}
a^2b^2 &= (k-9)(k-16)
\tag{6}\label{6}
.
\end{align}
Combination of \eqref{5} and \eqref{6}
results in equation in $k$:
\begin{align}
k(3k-52)&=0
\tag{7}\label{7}
,
\end{align}
and since $k>0$, we have the only option
\begin{align}
k&=\frac{52}3
\tag{8}\label{8}
.
\end{align}
Then \eqref{1} and \eqref{2} gives possible values of $a$
and $b$:
\begin{align}
a&=\pm \tfrac53\sqrt3
\tag{9}\label{9}
,\\
b&=\pm \tfrac23\sqrt3
\tag{10}\label{10}
.
\end{align}
And since $a,b$ also must be positive, the solution is ready.
