document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=212048>Question 212048</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0495117935/algebrahomeworkh></A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> The height of a triangle is 5 centimeters longer than three times its base. Find the base of the triangle if its area is 6 square centimeters </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #160139 by </B><A HREF=/tutors/aboutme.mpl?userid=drj><B>drj(1380)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=drj><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/cgi-bin/show-face.mpl?user=drj><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=160139\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The height of a triangle is 5 centimeters longer than three times its base. Find the base of the triangle if its area is 6 square centimeters\r<BR>\n" );
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document.write( "Step 1. Area of Triangle = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+xh%2F2+=6 ALIGN=MIDDLE  ALT=\"A+=+xh%2F2+=6\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "where x=base,  h=height and  A=area of is one-half times base times height\r<BR>\n" );
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document.write( "Step 2.  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h=3x%2B5 ALIGN=MIDDLE  ALT=\"h=3x%2B5\"   DISABLED_style= \"max-width:90%; height:auto;\" > (The height of a triangle is 5 centimeters longer than three times its base)\r<BR>\n" );
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document.write( "Step 3. Substitute height in Step 2 into equation in Step 1.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=xh%2F2=x%283%2Bx5%29%2F2 ALIGN=MIDDLE  ALT=\"A=xh%2F2=x%283%2Bx5%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%283x%5E2%2B5x%29%2F2=6 ALIGN=MIDDLE  ALT=\"%283x%5E2%2B5x%29%2F2=6\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Step 4.  Multiply 2 in both sides of last equation in Step 3 to get rid of denominator.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%283x%5E2%2B5x%29%2F2=2%2A6 ALIGN=MIDDLE  ALT=\"2%283x%5E2%2B5x%29%2F2=2%2A6\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2B5x=12 ALIGN=MIDDLE  ALT=\"3x%5E2%2B5x=12\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Simplify by subtracting 12 from both sides of equation yields\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2B5x-12=0 ALIGN=MIDDLE  ALT=\"3x%5E2%2B5x-12=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Step 5.  The last equation of Step 4 is simply a quadratic equation.  Then use the quadratic formula:\r<BR>\n" );
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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;\" >\r<BR>\n" );
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document.write( "where a=3, b=5, and c=-12\r<BR>\n" );
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document.write( " <TR><TD  ALIGN=MIDDLE>Solved by <A  rel=nofollow HREF=http://wiki.algebra.com/index.php/Tutoring_-_pluggable_solvers>pluggable</A> solver: <A  rel=nofollow HREF=quadratic.solver>SOLVE quadratic equation with variable</A></TD></TR>\n" );
document.write( "<TR><TD  ALIGN=LEFT>Quadratic equation <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ax%5E2%2Bbx%2Bc=0 ALIGN=MIDDLE  ALT=\"ax%5E2%2Bbx%2Bc=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > (in our case <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2B5x%2B-12+=+0 ALIGN=MIDDLE  ALT=\"3x%5E2%2B5x%2B-12+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >) has the following solutons:<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "  For these solutions to exist, the <U>discriminant</U> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac ALIGN=MIDDLE  ALT=\"b%5E2-4ac\"   DISABLED_style= \"max-width:90%; height:auto;\" > should not be a negative number.<BR>\n" );
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document.write( "  First, we need to compute the discriminant <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac ALIGN=MIDDLE  ALT=\"b%5E2-4ac\"   DISABLED_style= \"max-width:90%; height:auto;\" >: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac=%285%29%5E2-4%2A3%2A-12=169 ALIGN=MIDDLE  ALT=\"b%5E2-4ac=%285%29%5E2-4%2A3%2A-12=169\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
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document.write( "  Discriminant d=169 is greater than zero. That means that there are two solutions: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5B12%5D+=+%28-5%2B-sqrt%28+169+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"+x%5B12%5D+=+%28-5%2B-sqrt%28+169+%29%29%2F2%5Ca\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "    <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B1%5D+=+%28-%285%29%2Bsqrt%28+169+%29%29%2F2%5C3+=+1.33333333333333 ALIGN=MIDDLE  ALT=\"x%5B1%5D+=+%28-%285%29%2Bsqrt%28+169+%29%29%2F2%5C3+=+1.33333333333333\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "    <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B2%5D+=+%28-%285%29-sqrt%28+169+%29%29%2F2%5C3+=+-3 ALIGN=MIDDLE  ALT=\"x%5B2%5D+=+%28-%285%29-sqrt%28+169+%29%29%2F2%5C3+=+-3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "    Quadratic expression <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2B5x%2B-12 ALIGN=MIDDLE  ALT=\"3x%5E2%2B5x%2B-12\"   DISABLED_style= \"max-width:90%; height:auto;\" > can be factored:<BR>\n" );
document.write( "  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2B5x%2B-12+=+3%28x-1.33333333333333%29%2A%28x--3%29 ALIGN=MIDDLE  ALT=\"3x%5E2%2B5x%2B-12+=+3%28x-1.33333333333333%29%2A%28x--3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "  Again, the answer is: 1.33333333333333, -3.\n" );
document.write( "Here's your graph:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B5%2Ax%2B-12+%29 ALIGN=MIDDLE  ALT=\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B5%2Ax%2B-12+%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ></TD></TR>\n" );
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document.write( "Step 6.  Pick the positive number in the quadratic equation.  In this case, we have the base x= 1.3333=4/3.  \r<BR>\n" );
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document.write( "height is then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h=3x%2B5=3%284%2F3%29%2B5=9 ALIGN=MIDDLE  ALT=\"h=3x%2B5=3%284%2F3%29%2B5=9\"   DISABLED_style= \"max-width:90%; height:auto;\" > centimeters\r<BR>\n" );
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document.write( "Area is then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29%284%2F3%29%289%29=6 ALIGN=MIDDLE  ALT=\"%281%2F2%29%284%2F3%29%289%29=6\"   DISABLED_style= \"max-width:90%; height:auto;\" > square centimeters.  This is the same answer to our given answer.  So it check out.\r<BR>\n" );
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document.write( "For Step-By-Step Videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra.  For other technical topics having Step-By-Step videos in Trigonometry, Differential Equations, and Circuit Analysis and Design 1, please visit http://FreedomUniversity.TV.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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