document.write( "Question 30331: Quesion: Which wquation below is the quadratic equation in standard form that results whsn you substitute u+x^-2 into the equation x^-4 -32 = 4x^-2 ?\r
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\n" ); document.write( "(a) 1/u^2 - 4/u -32 =0
\n" ); document.write( "(b) u^2 -4u -32 = 0
\n" ); document.write( "(c) u^4 -4u^2 -32 =0
\n" ); document.write( "(d) u^-2 -4u +32 = 0
\n" ); document.write( "(e) 1/u^2 +4/u -32 =o
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Algebra.Com's Answer #17021 by Cintchr(481)\"\" \"About 
You can put this solution on YOUR website!
\"+x%5E-4+-32+=+4x%5E-2+\" and \"+u%2Bx%5E-2+\"
\n" ); document.write( "What are we substituting u+x^-2 for? What is it equal to? I am going to assume that \"+u%2Bx%5E-2+\" should actually be \"+u=x%5E-2+\"
\n" ); document.write( "First, set the first equation equal to 0
\n" ); document.write( "\"+x%5E-4+-32+-4x%5E-2+=+0+\" now ... the x^-4 needs to be rewritten
\n" ); document.write( "\"+%28x%5E-2%29%5E2+-32+-4x%5E-2+=+0+\" anywhere there is a x^-2 ... plug in a u
\n" ); document.write( "\"+u%5E2+-32+-4u+=+0+\" put the equation in descending order
\n" ); document.write( "\"+u%5E2-4u-32+=+0+\" the answer is b
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