Question 204243
since your equation is all strung out without parentheses to show the order of the arithmetic, it's difficult to see what's going on but i'll take a stab at it.
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it looks like it might be the following:
{{{x^(3/2)-4x^(3/4)+3=0}}}
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that would be written as follows without the fancy algebra.com formula generator.
x^(3/2)-4x^(3/4)+3=0
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now {{{x^(3/2) = (x^(3/4))^2}}}
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if we let {{{y}}} = {{{x^(3/4)}}} then {{{y^2}}} = {{{x^(3/2)}}} and our equation becomes:
{{{y^2 -4y +3 = 0}}} which is equivalent to:
{{{(y-3)*(y-1) = 0}}}
which means that:
y = 3
or
y = 1
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if y = 1, then:
{{{x^(3/4)}}} = {{{1}}}
and
{{{x^(3/2)}}} = {{{1^2}}}
solving for x gets {{{x = 1}}} in both cases.
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if y = 3, then:
{{{x^(3/4)}}} = {{{3}}}
and
{{{x^(3/2)}}} = {{{3^2}}}
solving for x gets {{{x = 3*root(3,3)}}}
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so you have an answer i think.
x = {{{1}}}
or 
x = {{{3*root(3,3)}}}
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we have to substitute in the original equation to see if they're good.
your original equation was:
{{{x^(3/2)-4x^(3/4)+3=0}}}
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when x = {{{1}}} this becomes:
{{{1^(3/2)-4*1^(3/4)+3=0}}}
which becomes:
{{{1 - 4 + 3 = 0}}}
which becomes:
{{{0 = 0}}}
so we know that x = 1 is good.
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{{{3*root(3,x)}}} is equivalent to {{{(3*3^(1/3))}}}
so when x = {{{3*root(3,x)}}} and we apply the substitute of x = {{{3*3^(1/3)}}} then your equation of:
{{{x^(3/2)-4x^(3/4)+3=0}}} becomes:
{{{(3*3^(1/3))^(3/2)-4(3*3^(1/3))^(3/4)+3=0}}}
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after all the arithmetic which i won't detail because i'm getting tired of typing formulas in algebra.com for the day, you get:
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{{{(3*3^(1/3))^(3/2)-4(3*3^(1/3))^(3/4)+3=0}}} becomes:
{{{3^2 - 4*(3) + 3 = 0}}} which becomes:
{{{9 - 12 + 3 = 0}}} which becomes
{{{0 = 0}}}
so we know that x = 3 is also good.
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the first part of the equation above was:
{{{(3*3^(1/3))^(3/2)}}}
i'll solve this part for you.
you can solve the other part if you're intere4sted.
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{{{(3*3^(1/3))^(3/2)}}} becomes:
{{{(3^(3/2) * (3^(1/3))^(3/2))}}} becomes:
{{{(3^(3/2) * (3^((1/3)*(3/2))))}}} becomes:
{{{(3^(3/2) * (3^(3/6)))}}} becomes:
{{{3^((3/2) + (3/6))}}} becomes:
{{{3^((9/6) + (3/6))}}} becomes:
{{{3^(12/6)}}} becomes:
{{{3^(2)}}} becomes:
{{{9}}}
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the answers to your problem are:
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{{{x = 1}}}
or 
{{{x = 3*root(3,3)}}}
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the equation i solved is:
{{{x^(3/2)-4x^(3/4)+3=0}}}
which would be written as follows without the fancy algebra.com formula generator.
x^(3/2)-4x^(3/4)+3=0
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i solved by seeing that:
{{{x^(3/2)}}} was equal to {{{(x^(3/4))^2}}}
so i let {{{y = x^(3/4)}}}
and i let {{{y^2 = x^(3/2)}}}
and i solved the quadratic equation:
{{{y^2 - 4y + 3}}}
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i'm done for the day.
this was interesting, to say the least.
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