what is $a^2+9=b^2+16=1+(a+b)^2$ solve for $a,b$

Question

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)

Answer 1

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?

Answer 2

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 $(a, b)=(5/\sqrt{3})$.

Answer 3

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)

Answer 4

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.