document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=3081>Question 3081</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find a polynomial that represents the area of a circle whose radius is b+1 meters.  Using the value of pi=3.14, what would be the radius if the area was to equate 4pi square meters? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #1312 by </B><A HREF=/tutors/aboutme.mpl?userid=thechamp1011><B>thechamp1011(19)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=thechamp1011><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=1312\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> WE KNOW THE FORMULA FOR AREA OF THE CIRCLE IS: A=Pi (r)^2<BR>\n" );
document.write( "if r =b+1......................................A=pi(b+1)^2<BR>\n" );
document.write( "                                               A=pi (b^2+2b+1)\r<BR>\n" );
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document.write( "FOR THE SECOND PART OF THE PROBLEM: A= PI (R)^2<BR>\n" );
document.write( "SINCE A=4PI.........................4PI=PI (R)^2<BR>\n" );
document.write( "DIVIDE BOTH SIDE BY PI..............4PI/PI=PI/PI (R)^2<BR>\n" );
document.write( "....................................4=(r)^2<BR>\n" );
document.write( "TO FIND R, TAKE SQRT ON BOTH SIDE....r=+-2 </SPAN>\n" );
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