SOLUTION: Solve the equation by making an appropriate substitution. x-512-16sqrtx=0 here is what I have tried: x^1-16x^1/2-512 (x^1/2)^2-16^1/2-512 subsitute x for u u^2-16u-512 facto

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Solve the equation by making an appropriate substitution. x-512-16sqrtx=0 here is what I have tried: x^1-16x^1/2-512 (x^1/2)^2-16^1/2-512 subsitute x for u u^2-16u-512 facto      Log On


   

Question 542571: Solve the equation by making an appropriate substitution.
x-512-16sqrtx=0
here is what I have tried:
x^1-16x^1/2-512
(x^1/2)^2-16^1/2-512
subsitute x for u
u^2-16u-512
factor (I cannot find a factor for this problem so I divide everything by 2)
u^2-8u-256
using the formula
8 +-sqrt8^2-4(1)(256)/2(1)
8 +-sqrrt1088 / 2
reduced
4+12sqrrt17, 4-12sqrrt17
my answer choices are
{768}
{2068}
{1024}
{512}
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You're ok up the point of substituting u+=+x%5E%281%2F2%29 or u+=+sqrt%28x%29
u%5E2-16u-512+=+0 This wil factor:
%28u%2B16%29%28u-32%29+=+0 Apply the zero product rule.
u%2B16+=+0 or u-32+=+0 so...
u+=+-16 or u+=+32 Substitute u+=+sqrt%28x%29
sqrt%28x%29+=+-16 or sqrt%28x%29+=+32 Square both sides in each case.
highlight%28x+=+256%29 or highlight_green%28x+=+1024%29
You should check these solutions because, in squaring, you may have inadvertently introduced invalid solutions.
x-512-16sqrt%28x%29+=+0 Substitute x+=+256
256-512-16sqrt%28256%29+=+0
256-512-16%2A16+=+0
256-512-256+=+0
-512+%3C%3E+0
1024-512-16sqrt%281024%29+=+0
1024-512-16%2A32+=+0
1024-512-512+=+0
1024-1024+=+0 OK