What is the speed of an electron when its kinetic energy is equal to 71 % of its total energy?

Question

The equations I tried to use are 1/2mv^2=.71ymc^2
y=1/sqroot (1-B^2) B=v/c Then I get stuck in the math, does anyone see a problem? maybe I I using the wrong equations to figure this out? All help is appreciated. Thanks

Meg I have all that but i get an answer greater than the speed of light and thats not possible care he help me through it?

I figured it out usuing (y-1)mc^2=.71ymc^2... so y-1=.71y and going from there, I am sure meg was saying the same thing I am just super slow, lol.

Answer 1

Total energy = KE + mc²= mc²/sqrt(1-v²/c²)
sqrt(1-v²/c²)={ mc²/(KE + mc²) }=1-.71