document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=82008>Question 82008</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Here is my problem: The hypotenuse of a right triangle is 1cm more than twice the shorter leg, and the longer leg is 9cm less than three times the shorter leg. Find the lengths of the three sides of the triangle. \r<BR>\n" );
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document.write( "I know that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5E2+%2B+b%5E2+=+c%5E2 ALIGN=MIDDLE  ALT=\"a%5E2+%2B+b%5E2+=+c%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > (pythagorean theorem), so if a is the shorter leg, I would get: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5E2+%2B+%283a-9%29%5E2+=+%282a%2B1%29%5E2 ALIGN=MIDDLE  ALT=\"a%5E2+%2B+%283a-9%29%5E2+=+%282a%2B1%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > So if I follow what I think I know about solving equations, I wind up with <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10a%5E2+-+54a+%2B+81+=+4a%5E2+%2B+4a+%2B1 ALIGN=MIDDLE  ALT=\"10a%5E2+-+54a+%2B+81+=+4a%5E2+%2B+4a+%2B1\"   DISABLED_style= \"max-width:90%; height:auto;\" > which I can simplify down to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6a%5E2+-+58a+=+-80 ALIGN=MIDDLE  ALT=\"6a%5E2+-+58a+=+-80\"   DISABLED_style= \"max-width:90%; height:auto;\" >. This is where I get stuck. I'm not sure how to solve for a with <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5E2 ALIGN=MIDDLE  ALT=\"a%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > in the equation, and if I try to factor out the a I am getting a negitive, never ending number for a.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #58753 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley75><B>checkley75(3666)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley75><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=58753\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A^2+(3A-9)^2=(2A+1)^2<BR>\n" );
document.write( "A^2+9A^2-54A+81=4A^2+4A+1<BR>\n" );
document.write( "10A^2-4A^2-54A-4A+81-1=0<BR>\n" );
document.write( "6A^2-58A+80=0 HERE YOU HAVE TO FIND FACTORS OF 6 & 80 WHEN MULTIPLIED & THEN ADDED WILL EQUAL -58.<BR>\n" );
document.write( "IF YOU CAN'T DO THAT THEN USE THE QUADRATIC EQUATION (aX^2+bX+c=0: <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ ALIGN=MIDDLE  ALT=\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <BR>\n" );
document.write( "(2A-16)(3A-5)=0<BR>\n" );
document.write( "2A-16=0<BR>\n" );
document.write( "2A=16<BR>\n" );
document.write( "A=16/2<BR>\n" );
document.write( "A=8 ANSWER.<BR>\n" );
document.write( "3A-5=0<BR>\n" );
document.write( "3X=5<BR>\n" );
document.write( "A=5/3<BR>\n" );
document.write( "A=1 2/3 ANSWER. </SPAN>\n" );
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