SOLUTION: How do you solve x3/2-4x3/4+3=0, I don't know where to start?

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Question 204243: How do you solve x3/2-4x3/4+3=0, I don't know where to start?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
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%5E%283%2F2%29-4x%5E%283%2F4%29%2B3=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%5E%283%2F2%29+=+%28x%5E%283%2F4%29%29%5E2
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if we let y = x%5E%283%2F4%29 then y%5E2 = x%5E%283%2F2%29 and our equation becomes:
y%5E2+-4y+%2B3+=+0 which is equivalent to:
%28y-3%29%2A%28y-1%29+=+0
which means that:
y = 3
or
y = 1
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if y = 1, then:
x%5E%283%2F4%29 = 1
and
x%5E%283%2F2%29 = 1%5E2
solving for x gets x+=+1 in both cases.
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if y = 3, then:
x%5E%283%2F4%29 = 3
and
x%5E%283%2F2%29 = 3%5E2
solving for x gets x+=+3%2Aroot%283%2C3%29
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so you have an answer i think.
x = 1
or
x = 3%2Aroot%283%2C3%29
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we have to substitute in the original equation to see if they're good.
your original equation was:
x%5E%283%2F2%29-4x%5E%283%2F4%29%2B3=0
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when x = 1 this becomes:
1%5E%283%2F2%29-4%2A1%5E%283%2F4%29%2B3=0
which becomes:
1+-+4+%2B+3+=+0
which becomes:
0+=+0
so we know that x = 1 is good.
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3%2Aroot%283%2Cx%29 is equivalent to %283%2A3%5E%281%2F3%29%29
so when x = 3%2Aroot%283%2Cx%29 and we apply the substitute of x = 3%2A3%5E%281%2F3%29 then your equation of:
x%5E%283%2F2%29-4x%5E%283%2F4%29%2B3=0 becomes:
%283%2A3%5E%281%2F3%29%29%5E%283%2F2%29-4%283%2A3%5E%281%2F3%29%29%5E%283%2F4%29%2B3=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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%283%2A3%5E%281%2F3%29%29%5E%283%2F2%29-4%283%2A3%5E%281%2F3%29%29%5E%283%2F4%29%2B3=0 becomes:
3%5E2+-+4%2A%283%29+%2B+3+=+0 which becomes:
9+-+12+%2B+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:
%283%2A3%5E%281%2F3%29%29%5E%283%2F2%29
i'll solve this part for you.
you can solve the other part if you're intere4sted.
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%283%2A3%5E%281%2F3%29%29%5E%283%2F2%29 becomes:
%283%5E%283%2F2%29+%2A+%283%5E%281%2F3%29%29%5E%283%2F2%29%29 becomes:
%283%5E%283%2F2%29+%2A+%283%5E%28%281%2F3%29%2A%283%2F2%29%29%29%29 becomes:
%283%5E%283%2F2%29+%2A+%283%5E%283%2F6%29%29%29 becomes:
3%5E%28%283%2F2%29+%2B+%283%2F6%29%29 becomes:
3%5E%28%289%2F6%29+%2B+%283%2F6%29%29 becomes:
3%5E%2812%2F6%29 becomes:
3%5E%282%29 becomes:
9
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the answers to your problem are:
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x+=+1
or
x+=+3%2Aroot%283%2C3%29
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the equation i solved is:
x%5E%283%2F2%29-4x%5E%283%2F4%29%2B3=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%5E%283%2F2%29 was equal to %28x%5E%283%2F4%29%29%5E2
so i let y+=+x%5E%283%2F4%29
and i let y%5E2+=+x%5E%283%2F2%29
and i solved the quadratic equation:
y%5E2+-+4y+%2B+3
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i'm done for the day.
this was interesting, to say the least.
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