Question 1199216
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The {{{3x^(1/2)^""}}} is not the same as {{{(3x)^(1/2)^""}}}
In the first expression, we apply the 1/2 exponent to only the x; while the second expression has the exponent apply to all of 3x.


Recall that with PEMDAS, the parenthesis is done first, then exponents, then multiplication.
P = parenthesis
E = exponents
M = multiplication
D = division
A = addition
S = subtraction


The reason I bring this up is that with {{{(3x)^(1/2)^""}}}, we compute the 3x first in the parenthesis, then apply the exponent later.
Whereas with {{{3x^(1/2)^""}}}, we do the exponent piece {{{x^(1/2)^""}}} first before multiplying with the 3 out front.


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Here is one way to solve for x.


{{{x^(3/2)^"" - 3x^(1/2)^"" = 0}}}


{{{(x^3)^(1/2)^"" - 3x^(1/2)^"" = 0}}} Rearrange exponents in the first term. Use the rule a^(b*c) = (a^b)^c.


{{{sqrt(x^3) - 3*sqrt(x) = 0}}} Raising to the 1/2 power is the same as a square root.


{{{sqrt(x^3) = 3*sqrt(x)}}}


{{{(sqrt(x^3))^2 = (3*sqrt(x))^2}}} Square both sides


{{{x^3 = 9x}}} 


{{{x^3-9x = 0}}}


{{{x(x^2-9) = 0}}}


{{{x(x-3)(x+3)=0}}}


{{{x = 0}}} or {{{x-3=0}}} or {{{x+3=0}}}


{{{x = 0}}} or {{{x=3}}} or {{{x=-3}}}
Those are the potential solutions. 
But we need to check each of them.


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If you were to plug in x = 0 into the original equation, then,
{{{x^(3/2)^"" - 3x^(1/2)^"" = 0}}}


{{{sqrt(x^3) - 3*sqrt(x) = 0}}}


{{{sqrt(0^3) - 3*sqrt(0) = 0}}}


{{{0 - 3*0 = 0}}}


{{{0 - 0 = 0}}}


{{{0 = 0}}}
We end up with a true statement, therefore confirming x = 0 is indeed a root.


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Repeat for x = 3
{{{x^(3/2)^"" - 3x^(1/2)^"" = 0}}}


{{{sqrt(x^3) - 3*sqrt(x) = 0}}}


{{{sqrt(3^3) - 3*sqrt(3) = 0}}}


{{{sqrt(3^2*3) - 3*sqrt(3) = 0}}}


{{{sqrt(3^2)*sqrt(3) - 3*sqrt(3) = 0}}}


{{{3*sqrt(3) - 3*sqrt(3) = 0}}}


{{{0 = 0}}}
We have confirmed x = 3 is also a root.


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Now try x = -3
{{{x^(3/2)^"" - 3x^(1/2)^"" = 0}}}


{{{sqrt(x^3) - 3*sqrt(x) = 0}}}


{{{sqrt((-3)^3) - 3*sqrt(-3) = 0}}}
We see that a negative is under the square root, to produce a complex value in the form a+bi, where {{{i = sqrt(-1)}}}
So we'll ignore x = -3.


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Another approach to take is to use a factoring method similar to what @greenestamps mentioned.
{{{x^(3/2)^"" - 3x^(1/2)^"" = 0}}}


{{{sqrt(x^3) - 3*sqrt(x) = 0}}}


{{{sqrt(x^2*x) - 3*sqrt(x) = 0}}}


{{{sqrt(x^2)*sqrt(x) - 3*sqrt(x) = 0}}}


{{{x*sqrt(x) - 3*sqrt(x) = 0}}}


{{{(x-3)*sqrt(x) = 0}}}


{{{x-3=0}}} or {{{sqrt(x)=0}}}


{{{x = 3}}} or {{{x = 0}}}
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