document.write( "Question 217614: {{ x^1/2-3x^1/4+2=0 }} \n" ); document.write( "

Algebra.Com's Answer #164074 by jsmallt9(3758) $\"\"$ $\"About$

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$\"+x%5E%281%2F2%29-3x%5E%281%2F4%29%2B2=0+\"$
\n" ); document.write( "There are a couple of keys to a relatively simple solution to this problem:

Recognize that 1/2 is two times 1/4
Recognize that because 1/2 is 2 times 1/4 and because of the rules for exponents, $\"x%5E%281%2F2%29+=+%28x%5E%281%2F4%29%29%5E2\"$

\n" ); document.write( "Once you get that $\"x%5E%281%2F2%29+=+%28x%5E%281%2F4%29%29%5E2\"$ , the rest of the problem is fairly simple. Without seeing this, the problem would be very difficult.

\n" ); document.write( "Your equation is of the form $\"q%5E2+-+3q+%2B2+=+0\"$ (with $\"q+=+x%5E%281%2F4%29\"$ ). And we solve this by factoring (or with the quadratic formula):
\n" ); document.write( " $\"%28q-2%29%28q-1%29=0\"$
\n" ); document.write( "The only way for this product (multiplication) to be zero is if one of the factors is zero. So we simply state this with equations:
\n" ); document.write( "q-2 = 0 or q-1 = 0
\n" ); document.write( "Solving each we get:
\n" ); document.write( "q = 2 or q = 1
\n" ); document.write( "Of course we're not interested in \"q\". So we substitute back in for q:
\n" ); document.write( " $\"x%5E%281%2F4%29+=+2\"$ or $\"x%5E%281%2F4%29+=+1\"$
\n" ); document.write( "Raising each side of each equation to the 4th power:
\n" ); document.write( "x = 16 or x = 1

\n" ); document.write( "Once you've done a few of these you will not need a substitute variable. You'll see that $\"+x%5E%281%2F2%29-3x%5E%281%2F4%29%2B2=0+\"$ will factor into $\"%28x%5E%281%2F4%29-2%29%28x%5E%281%2F4%29-1%29+=+0\"$ . And once you have a product equal to zero you are pretty close to the final solution. \n" ); document.write( "

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