document.write( "Question 1097197: Solve the following equations:\r
\n" ); document.write( "\n" ); document.write( "1.) X= (2/x+1)
\n" ); document.write( "2.) x+3 =(28/x)
\n" ); document.write( "3.) (3/x+1)=sq. root of (x+2)
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Algebra.Com's Answer #711606 by Boreal(15235)\"\" \"About 
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x=(2/(x+1))
\n" ); document.write( "x(x+1)=2
\n" ); document.write( "x^2+x-2=0
\n" ); document.write( "(x+2)(x-1)=0
\n" ); document.write( "x=-2, 1
\n" ); document.write( "check in original
\n" ); document.write( "-2=2/(-1), checks
\n" ); document.write( "1=2/2, checks.
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\n" ); document.write( "x+3=(28/x)
\n" ); document.write( "x^2+3x=28
\n" ); document.write( "x^2+3x-28=0
\n" ); document.write( "(x+7)(x-4)=0
\n" ); document.write( "x=4, -7
\n" ); document.write( "checks
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\n" ); document.write( "3/(x+1)=sqrt(x+2)
\n" ); document.write( "9/(x^2+2x+1)=x+2
\n" ); document.write( "9=x^3+2x^2+2x^2+4x+x+2
\n" ); document.write( "0=x^3+4x^2+5x-7
\n" ); document.write( "numerically, this is x=0.79, which when substituted has 1.675=1.67, same given rounding.
\n" ); document.write( "\"graph%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E3%2B4x%5E2%2B5x-7%29\"
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