SOLUTION: I am stumped. Please help me solve this equation... {{{(z/4-5)^2/3 = 1/25}}} I'm not sure if I am supposed to multiply the left expression by a power of ^3/2 to get rid of the p

Algebra ->  Radicals -> SOLUTION: I am stumped. Please help me solve this equation... {{{(z/4-5)^2/3 = 1/25}}} I'm not sure if I am supposed to multiply the left expression by a power of ^3/2 to get rid of the p      Log On


   

Question 535479: I am stumped. Please help me solve this equation...
%28z%2F4-5%29%5E2%2F3+=+1%2F25
I'm not sure if I am supposed to multiply the left expression by a power of ^3/2 to get rid of the power and then also do the same to the right.I know the left expression can be rewritten as the third root of z/4-5 to the second power.
Found 2 solutions by Alan3354, josmiceli:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
%28z%2F4-5%29%5E2%2F3+=+1%2F25
Multiply by 3
%28z%2F4-5%29%5E2+=+3%2F25
Take the sqrt
z%2F4+-+5+=+sqrt%283%29%2F5
Multiply by 4
z+-+20+=+4sqrt%283%29%2F5
z+=+20+%2B+4sqrt%283%29%2F5

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
%28z%2F4-5%29%5E2%2F3+=+1%2F25
I would change +1%2F25 to +%281%2F5%29%5E2+
%28z%2F4-5%29%5E2%2F3+=+%281%2F5%29%5E2
Then I would change +1%2F3 to +1%2F%28sqrt%283%29%29%5E2+
Now I have
+%28+%28z%2F4-5%29%2Fsqrt%283%29+%29%5E2+=+%281%2F5%29%5E2
Take the square root of both sides
++%28z%2F4-5%29%2Fsqrt%283%29++=+1%2F5
+z%2F4+-+5+=+sqrt%283%29%2F5+
+z%2F4+=+sqrt%283%29%2F5+%2B+5+
+z+=+%284%2F5%29%2Asqrt%283%29+%2B+20+
check answer:
+z%2F4+=+%281%2F5%29%2Asqrt%283%29+%2B+5+
+z%2F4+-+5+=+%281%2F5%29%2Asqrt%283%29+
+%28+z%2F4+-+5+%29%5E2+=+3%2F25+
+%28%28+z%2F4+-+5+%29%5E2%29%2F3+=+1%2F25+
OK