SOLUTION: Need help! I have to turn in this assignment by tonight and I'm having problems with this. 1) Solve the following equations. a) sqrt(x) - 2=1 Answer: Show work in th

Algebra ->  Algebra  -> Radicals -> SOLUTION: Need help! I have to turn in this assignment by tonight and I'm having problems with this. 1) Solve the following equations. a) sqrt(x) - 2=1 Answer: Show work in th      Log On

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Question 37643: Need help! I have to turn in this assignment by tonight and I'm having problems with this.
1) Solve the following equations.
a) sqrt(x) - 2=1
Answer:
Show work in this space.



b)sqrt(x)^3 = 27
Answer:
Show work in this space.



c)3^sqrt(x)^2 = 9
Answer:
Show work in this space.

Found 2 solutions by longjonsilver, stanbon:
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
1. +sqrt%28x%29+-+2+=+1+
+sqrt%28x%29+=+3+
now square both sides:
x = 9

2. +sqrt%28x%5E3%29+=+27+
well you would square both sides then take the cube root of both sides. Alternatively since "raising to a power" and "rooting" are opposite processes, it doesn't actually matter the order we do this, so i shall find the cube root first:

+sqrt%28x%29+=+root%283%2C+27%29+
+sqrt%28x%29+=+3+
--> x = 9
3. +3%5E%28sqrt%28x%29%5E2%29+=+9+
or... +3%5E%28sqrt%28x%29%5E2%29+=+3%5E2+
so by "symmetry" of this, we can say that +sqrt%28x%5E2%29+=+2+
and also, sqrt%28x%5E2%29+ is just x anyway, so
x = 2

jon.

Answer by stanbon(48535) About Me  (Show Source):
You can put this solution on YOUR website!
a) sqrt(x) - 2=1
sqrt(x)=3
Square both sides to get:
x=9

b)sqrt(x)^3 = 27
Square both sides to get:
x^3=27^2
Take the cube root of both sides to get:
x=(27^2)^(1/3)=[27^(1/3]^2
x=3^2=9

c)3^sqrt(x)^2 = 9
3^(sqrt(x)^2)= 3^2
sqrt(x)^2=2
Square both sides to get:
x^2=4
x=2
Cheers,
Stan H.