Question 1129801
let the roots be r and t, with r = 2t<br>

Then we can say:
{{{  r^2 - mr + 2 = 0 }}}   (1)
{{{  t^2 - mt + 2 = 0 }}}   (2)<br>

Substituting r=2t into (1): 
{{{  4t^2 - 2mt + 2 = 0 }}}<br>

This last equation and (2) can be solved for t in terms of m:
t = 3/m   ( —> r = 6/m )<br>

Plugging in t = 3/m  into (2):
{{{ (9/m^2) - m(3/m) + 2 = 0 }}}
{{{  9/m^2 = 1 }}}
{{{  m^2 =  9 }}}
      m = +/- 3 <br>

Both values of m work as solutions:
m=-3:
{{{ x^2 + 3x + 2 = 0 }}} —>  {{{ (x+1)(x+2) = 0 }}} —>  x = -1 and x = -2 <br>

m=3:
{{{ x^2 - 3x + 2 = 0 }}}  —>  {{{ (x-1)(x-2) = 0 }}}  —>  x = 1 and x = 2 <br>