Answer
$v = \sqrt{\frac{\sqrt{5}-1}{2}}\times c$
Work Step by Step
We can write an expression for the Newtonian value of the kinetic energy:
$K = \frac{1}{2}mv^2$
We can write an expression for the relativistic kinetic energy:
$K_r = (\gamma-1) ~mc^2$
If the kinetic energy is twice the Newtonian value, then:
$(\gamma-1) ~mc^2 = (2)(\frac{1}{2}mv^2)$
$(\gamma-1) ~mc^2 = mv^2$
$(\gamma-1) ~c^2 = v^2$
$\gamma = 1+\frac{v^2}{c^2}$
$\gamma = \frac{v^2+c^2}{c^2}$
We can find the speed when $\gamma = \frac{v^2+c^2}{c^2}$:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\gamma}$
$1-\frac{v^2}{c^2} = \frac{1}{\gamma^2}$
$\frac{c^2-v^2}{c^2} = \frac{c^4}{(v^2+c^2)^2}$
$(c^2-v^2)(v^2+c^2)^2 = c^6$
$(c^2-v^2)(v^4+2~v^2~c^2+c^4) = c^6$
$v^4c^2+2~v^2~c^4+c^6-v^6-2~v^4~c^2-v^2c^4 = c^6$
$-v^4c^2+v^2~c^4-v^6 = 0$
$v^2~(v^4+v^2c^2-c^4) = 0$
We can use the quadratic formula:
$v^2 = \frac{-c^2\pm \sqrt{(c^2)^2-(4)(1)(-c^4)}}{2(1)}$
$v^2 = \frac{-c^2\pm \sqrt{c^4+4c^4}}{2}$
$v^2 = \frac{-c^2\pm \sqrt{5c^4}}{2}$
$v^2 = \frac{-c^2\pm \sqrt{5}~c^2}{2}$
$v^2 = \frac{(-\sqrt{5}-1)~c^2}{2}, \frac{(\sqrt{5}-1)~c^2}{2}$
Since the speed is positive, the solution is $~~v^2 = \frac{(\sqrt{5}-1)~c^2}{2}$
Then:
$v = \sqrt{\frac{\sqrt{5}-1}{2}}\times c$
