SOLUTION: (1) sqrt of the fraction 27/16 (2) 3 plus the sqrt of x^2-8x=0 (3) (1 minus the sqrt of x)^2 How do I solve these? I tried the first one and got sqrt of the fraction

Algebra ->  Square-cubic-other-roots -> SOLUTION: (1) sqrt of the fraction 27/16 (2) 3 plus the sqrt of x^2-8x=0 (3) (1 minus the sqrt of x)^2 How do I solve these? I tried the first one and got sqrt of the fraction       Log On


   

Question 339116: (1) sqrt of the fraction 27/16
(2) 3 plus the sqrt of x^2-8x=0
(3) (1 minus the sqrt of x)^2

How do I solve these?
I tried the first one and got sqrt of the fraction 3/4. Is that right? If not, show me the right steps to get the correct result.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
1) Find:
sqrt%2827%2F16%29
sqrt%2827%29+=+3sqrt%283%29
sqrt%2816%29+=+4 so...
highlight%28sqrt%2827%2F16%29+=+3%2Asqrt%283%29%2F4%29%29
2) Solve:
3%2Bsqrt%28x%5E2-8x%29+=+0 First, subtract 3 from both sides.
sqrt%28x%5E2-8x%29+=+-3 Now square both sides.
x%5E2-8x+=+9 Next, subtract 9 from both sides.
x%5E2-8x-9+=+0 Factor this trinomial.
%28x%2B1%29%28x-9%29+=+0 Now apply the zero product rule.
x%2B1+=+0 or x-9+=+0 so...
highlight%28x+=+-1%29 or highlight%28x+=+9%29
3) Simplify:
%281-sqrt%28x%29%29%5E2
%281-sqrt%28x%29%29%2A%281-sqrt%28x%29%29+=+highlight%281-2%2Asqrt%28x%29%2Bx%29