Question 478840
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/x^2}}}
-2y = {{{3/x^2}}} - x^2
Divide both sides by -2
y = {{{-3/(2x^2)}}} + {{{x^2/2}}}
put over a single denominator
y = {{{(-3 + x^4)/(2x^2)}}}
y = {{{(x^4-3)/(2x^2)}}}
Solve for x
{{{(x^4-3)/(2x^2)}}} = 0
Divide both sides by 2x^2
x^4 - 3 = 0
x^4  = 3
x = 3^(1/4)
x =  +/- 1.316
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Find all the solutions algebraically
1) {{{sqrt(x) - sqrt(x-5)}}} = 1
Rearrange this to
{{{sqrt(x) - 1 = sqrt(x-5)}}}
Square both sides
{{{(sqrt(x)-1)^2)}}} = x - 5
FOIL the left side
x - 2({{{sqrt(x)}}} + 1 = x - 5
-2({{{sqrt(x)}}} = x -x - 5 - 1
-2({{{sqrt(x)}}} = -6
Square both sides again
4x = 36
x = {{{36/4}}}
x = 9
Check this in the original equation
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2){{{3x(x-1)^(1/2)+2(x-1)^(1/2)}}}=0
{{{3x(x-1)^(1/2)}}} = {{{-2(x-1)^(1/2)}}}
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 = {{{2/3}}}, sadly, this will work in the original equation, no solution for x