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problem out of book.. x^(1/2) -3x^(1/4) +2=0
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substitute\[u^{2} = \sqrt{x}\]and then you will get a quadratic equation
like \[x ^{2}-3x+2=0\]?
yes
so then \[x ^{2}-3x+2+0\] \[(x-2)(x+1)= 2 and 1?\]the book said the answer was 1 and 16..
\[u^{2} - 3u + 2=0\]\[\left( u - 2 \right)\left( u-1\right)=0\]\[u = 2\]or\[u = 1\]now you can find x by plug in u to the substitution
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