Question 36741
{{{ sqrt(x+1) = 1 - sqrt(2x) }}}


The aim is to get the square roots on one side so that we can square both sides to remove the square roots. As i say, that is the aim but in this question, we have to do some further work:


{{{ sqrt(x+1) + sqrt(2x) = 1 }}}
{{{ (sqrt(x+1) + sqrt(2x))^2 = 1^2 }}}
{{{ (sqrt(x+1) + sqrt(2x))*(sqrt(x+1) + sqrt(2x)) = 1 }}}
{{{ sqrt(x+1)sqrt(x+1) + sqrt(x+1)sqrt(2x) + sqrt(2x)sqrt(x+1) + sqrt(2x)sqrt(2x) = 1 }}}
{{{ (x+1) + sqrt(x+1)sqrt(2x) + sqrt(2x)sqrt(x+1) + 2x = 1 }}}
{{{ (x+1) + 2sqrt(x+1)sqrt(2x) + 2x = 1 }}}
{{{ 3x+1 + 2sqrt(x+1)sqrt(2x) = 1 }}}
{{{ 3x + 2sqrt(x+1)sqrt(2x) = 0 }}}
{{{ 3x + 2sqrt((x+1)*(2x)) = 0 }}}
{{{ 2sqrt((x+1)*(2x)) = -3x }}}
{{{ sqrt((x+1)*(2x)) = -3x/2 }}}


and now square both sides again:
{{{ (sqrt((x+1)*(2x)))^2 = (-3x/2)^2 }}}
{{{ ((x+1)*(2x)) = 9x^2/4 }}}
{{{ 4((x+1)*(2x)) = 9x^2 }}}
{{{ 4(2x^2+2x) = 9x^2 }}}
{{{ 8x^2+8x = 9x^2 }}}
{{{ 0 = x^2 - 8x }}}
{{{ x^2 - 8x = 0 }}}
x(x-8) = 0
so x=0 or x-8=0
--> x=0 or x=8


jon.