an equation is shown.
[the square root of (x^{2}-t^{2})] = 2t-x
If x and t are positive numbers that satisfy the equation above, what is the value of (x/t)?
$\displaystyle \sqrt{x^2-t^2}=2t-x$
$\displaystyle \Rightarrow x^2-t^2 = 4t^2-4tx+x^2$
$\displaystyle \Rightarrow 5t^2-4tx = 0$
$\displaystyle \Rightarrow t(5t - 4x) = 0$
$\displaystyle t > 0$, so
$\displaystyle 5t-4x = 0\quad\Rightarrow\quad4x=5t\quad\Rightarrow\quad \frac xt=\frac54$
Hint
$\displaystyle \begin{gathered} \sqrt {{x^2} - {t^2}} = 2t - x \hfill \\ t\sqrt {{{\left( {\frac{x}{t}} \right)}^2} - 1} = 2t - x \hfill \\ \sqrt {{{\left( {\frac{x}{t}} \right)}^2} - 1} = 2 - \frac{x}{t} \hfill \\ \text{Let}\,\frac{x}{t} = a,\quad\text{domain}\colon\,|a| \geqslant 1 \hfill \\ \sqrt {{a^2} - 1} = 2 - a \hfill \\ {a^2} - 1 = {(2 - a)^2} \hfill \\{a^2} - 1 = 4 - 4a + {a^2} \hfill \\ 4a = 5 \hfill \\ a = \frac{5}{4} \hfill \\ \end{gathered}$