Question 145996
{{{y + 6*sqrt(y) = 16}}}
I define a new variable to be {{{z^2 = y}}}
Taking the square root of both sides, {{{z =0 +-sqrt(y)}}}
Sustituting,
{{{z^2 + 6z -16 = 0}}}
{{{(z + 8)(z - 2) = 0}}}
{{{z = -8}}}
{{{z = 2}}}
Putting {{{z}}} back in terms of {{{y}}}
{{{-sqrt(y) = -8}}}
{{{y = 64}}}
and
{{{sqrt(y) = 2}}}
{{{y = 4}}}
Now I substitute these answers  in the original equation
{{{y + 6*sqrt(y) = 16}}}
{{{64 - 6*8 = 16}}}
{{{64 - 48 = 16}}}
{{{16 = 16}}}
and
{{{4 + 6*2 = 16}}}
{{{4 + 12 = 16}}}
It looks like they are only allowing {{{sqrt(y)}}} to be
positive, so {{{y = 4}}} and {{{sqrt(y) = 2}}} are the answers