Solve $a^2 + b^2 = t, 2ab = 1/t$ for $a,b$ I want to solve
$$
a^2 + b^2 = t \\
2ab = 1/t
$$
for $a$ and $b$ in terms of $t$. Wolfram solves it but the solution is very dirty.
 A: $$(a+b)^2 = a^2+b^2+2ab = t + \frac{1}{t}$$
so we have
$$
a + b = \pm\sqrt{t+\frac{1}{t}}
$$
and proceed.
A: $${a^2 + b^2 = t \\}$$
$${2ab = 1/t}$$
$${(a+b)^2}=t+1/t$$
$${a+b= \pm\sqrt{t+\frac{1}{t}}}$$
Also,
$${(a-b)^2=t-1/t}$$
$${a-b= \pm\sqrt{t-\frac{1}{t}}}$$
Just make some addition and subtraction now to get the answer. I shall let you conclude.
A: $$
\left \{ \begin{array}{rclr}
  a^2+b^2 &=& t & \qquad \cdots \cdots (1) \\
  2ab &=& \dfrac{1}{t} & \qquad \cdots \cdots (2) \\
\end{array}
\right.$$
From $(1)$, $$t \ge 0$$
By $AM \ge GM$, $$t\ge \frac{1}{t} \implies t\ge 1$$
$(1)+(2)$, $$(a+b)^2=t+\frac{1}{t} \implies |a+b|=\sqrt{t+\frac{1}{t}}$$
$(1)-(2)$, $$(a-b)^2=t-\frac{1}{t} \implies |a-b|=\sqrt{t-\frac{1}{t}}$$
Since $2ab=\dfrac{1}{t} \in (0,1]$,
$$
\begin{pmatrix} a \\ b \end{pmatrix}=
\begin{pmatrix}
  \frac{1}{2} \left( \sqrt{t+\frac{1}{t}} \pm \sqrt{t-\frac{1}{t}} \right) \\
  \frac{1}{2} \left( \sqrt{t+\frac{1}{t}} \mp \sqrt{t-\frac{1}{t}} \right)
\end{pmatrix}
\quad \text{or} \quad
\begin{pmatrix}
  \frac{1}{2} \left( -\sqrt{t+\frac{1}{t}} \pm \sqrt{t-\frac{1}{t}} \right) \\
  \frac{1}{2} \left( -\sqrt{t+\frac{1}{t}} \mp \sqrt{t-\frac{1}{t}} \right)
\end{pmatrix}$$

P.S.
The solution by Wolfram Alpha isn't "dirty", it takes care the sign of $2ab$ after simplification.
Eliminating $b$ gives biquadratic in $a$:
$$4t^2a^4-4t^3a^2+1=0 \implies a^2=\frac{t^2 \pm \sqrt{t^4-1}}{2t}$$
$$
\begin{pmatrix} a \\ b \end{pmatrix}=
\begin{pmatrix}
  \sqrt{\dfrac{t^2 \pm \sqrt{t^4-1}}{2t}} \\
  \sqrt{\dfrac{t^2 \mp \sqrt{t^4-1}}{2t}}
\end{pmatrix}
\quad \text{or} \quad
\begin{pmatrix}
  -\sqrt{\dfrac{t^2 \pm \sqrt{t^4-1}}{2t}} \\
  -\sqrt{\dfrac{t^2 \mp \sqrt{t^4-1}}{2t}}
\end{pmatrix}
$$

