Question 542571
You're ok up the point of substituting {{{u = x^(1/2)}}} or {{{u = sqrt(x)}}}
{{{u^2-16u-512 = 0}}} This wil factor:
{{{(u+16)(u-32) = 0}}} Apply the zero product rule.
{{{u+16 = 0}}} or {{{u-32 = 0}}} so...
{{{u = -16}}} or {{{u = 32}}} Substitute {{{u = sqrt(x)}}}
{{{sqrt(x) = -16}}} or {{{sqrt(x) = 32}}} Square both sides in each case.
{{{highlight(x = 256)}}} or {{{highlight_green(x = 1024)}}}
You should check these solutions because, in squaring, you may have inadvertently introduced invalid solutions.
{{{x-512-16sqrt(x) = 0}}} Substitute {{{x = 256}}}
{{{256-512-16sqrt(256) = 0}}}
{{{256-512-16*16 = 0}}}
{{{256-512-256 = 0}}}
{{{-512 <> 0}}}
{{{1024-512-16sqrt(1024) = 0}}}
{{{1024-512-16*32 = 0}}}
{{{1024-512-512 = 0}}}
{{{1024-1024 = 0}}} OK