SOLUTION: My question is: Solve this equation algebraically x^5=3x+2x^3y and: Find all the solutions algebraically 1) Sqroot(x) − Sqroot(x − 5) = 1 2)3x(x-1)^(1/2)+2

Algebra ->  Equations -> SOLUTION: My question is: Solve this equation algebraically x^5=3x+2x^3y and: Find all the solutions algebraically 1) Sqroot(x) − Sqroot(x − 5) = 1 2)3x(x-1)^(1/2)+2      Log On


   

Question 478840: My question is: Solve this equation algebraically
x^5=3x+2x^3y
and: Find all the solutions algebraically
1) Sqroot(x) − Sqroot(x − 5) = 1
2)3x(x-1)^(1/2)+2(x-1)^(1/2)=0

If you can please explain how to solve it for future reference. Thank You :)
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Solve this equation algebraically
x^5 = 3x + 2x^3y
x^5 - 2x^3y = 3x
Factor out x^3
x^3(x^2 - 2y) = 3x
divide both sides by x
x^2(x^2 - 2y) = 3
x^2 - 2y = 3%2Fx%5E2
-2y = 3%2Fx%5E2 - x^2
Divide both sides by -2
y = -3%2F%282x%5E2%29 + x%5E2%2F2
put over a single denominator
y = %28-3+%2B+x%5E4%29%2F%282x%5E2%29
y = %28x%5E4-3%29%2F%282x%5E2%29
Solve for x
%28x%5E4-3%29%2F%282x%5E2%29 = 0
Divide both sides by 2x^2
x^4 - 3 = 0
x^4 = 3
x = 3^(1/4)
x = +/- 1.316
:
:
Find all the solutions algebraically
1) sqrt%28x%29+-+sqrt%28x-5%29 = 1
Rearrange this to
sqrt%28x%29+-+1+=+sqrt%28x-5%29
Square both sides
%28sqrt%28x%29-1%29%5E2%29 = x - 5
FOIL the left side
x - 2(sqrt%28x%29 + 1 = x - 5
-2(sqrt%28x%29 = x -x - 5 - 1
-2(sqrt%28x%29 = -6
Square both sides again
4x = 36
x = 36%2F4
x = 9
Check this in the original equation
:
:
2)3x%28x-1%29%5E%281%2F2%29%2B2%28x-1%29%5E%281%2F2%29=0
3x%28x-1%29%5E%281%2F2%29 = -2%28x-1%29%5E%281%2F2%29
Square both sides
9x^2(x-1) = 4(x-1)
divide both sides by (x-1), leaving us with
9x^2 = 4
Find the square root of both sides
3x = 2
x = 2%2F3, sadly, this will work in the original equation, no solution for x