document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=845255>Question 845255</A>: <I><SPAN STYLE=\"word-break: break-all;\"> solve the equation for x. If neccesary, round your answer to two decimal places.<BR>\n" );
document.write( "x+2=square root of 3x+16 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #509200 by </B><A HREF=/tutors/aboutme.mpl?userid=pmesler><B>pmesler(52)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=pmesler><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=pmesler><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=509200\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The equation says x+2 = sqrt(3x+16) \r<BR>\n" );
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document.write( "First, to solve for x we need to simplify the equation. To do that we need to eliminate the square root symbol. We do that by performing the opposite or inverse function. The inverse function of a square root is to square something. It's important to note that we need to do this for both sides of the equation.\r<BR>\n" );
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document.write( "Therefore our new equation is \r<BR>\n" );
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document.write( "(x+2)^2 = (sqrt(3x+16))^2\r<BR>\n" );
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document.write( "That gives us\r<BR>\n" );
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document.write( "(x+2)(x+2) = 3x+16\r<BR>\n" );
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document.write( "Use the FOIL method to simplify the left-hand side.\r<BR>\n" );
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document.write( "x^2+4x+4 = 3x+16\r<BR>\n" );
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document.write( "Right away we see that our equation is becoming a quadratic equation. Let's simplify the whole equation so we have a quadratic expression on the left side and a zero on the right side.\r<BR>\n" );
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document.write( "Let's subtract 3x from both sides to get\r<BR>\n" );
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document.write( "x^2 + x + 4 = 16\r<BR>\n" );
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document.write( "Now, let's get rid of the 16 so we're left with a zero on the right-hand side.\r<BR>\n" );
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document.write( "Subtract 16 from both sides.\r<BR>\n" );
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document.write( "x^2+x-12 = 0.\r<BR>\n" );
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document.write( "Now, we use the quadratic formula to solve for x. \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=1x%5E2%2B1x%2B-12+=+0 ALIGN=MIDDLE  ALT=\"1x%5E2%2B1x%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" );
document.write( "  <BR>\n" );
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" );
document.write( "  <BR>\n" );
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=%281%29%5E2-4%2A1%2A-12=49 ALIGN=MIDDLE  ALT=\"b%5E2-4ac=%281%29%5E2-4%2A1%2A-12=49\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  Discriminant d=49 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+=+%28-1%2B-sqrt%28+49+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"+x%5B12%5D+=+%28-1%2B-sqrt%28+49+%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-%281%29%2Bsqrt%28+49+%29%29%2F2%5C1+=+3 ALIGN=MIDDLE  ALT=\"x%5B1%5D+=+%28-%281%29%2Bsqrt%28+49+%29%29%2F2%5C1+=+3\"   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-%281%29-sqrt%28+49+%29%29%2F2%5C1+=+-4 ALIGN=MIDDLE  ALT=\"x%5B2%5D+=+%28-%281%29-sqrt%28+49+%29%29%2F2%5C1+=+-4\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "    <BR>\n" );
document.write( "    Quadratic expression <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1x%5E2%2B1x%2B-12 ALIGN=MIDDLE  ALT=\"1x%5E2%2B1x%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=1x%5E2%2B1x%2B-12+=+1%28x-3%29%2A%28x--4%29 ALIGN=MIDDLE  ALT=\"1x%5E2%2B1x%2B-12+=+1%28x-3%29%2A%28x--4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "  Again, the answer is: 3, -4.\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+1%2Ax%5E2%2B1%2Ax%2B-12+%29 ALIGN=MIDDLE  ALT=\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-12+%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ></TD></TR>\n" );
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document.write( "The solutions are x = -4 and x = 3\r<BR>\n" );
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document.write( "Let's check our answers to see which of these roots is extraneous. \r<BR>\n" );
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document.write( "If we substitute x = -4 into the original equation we get this:\r<BR>\n" );
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document.write( "-4 + 2 = sqrt(3(-4) + 16) \r<BR>\n" );
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document.write( "-2 = sqrt(-12 + 16)\r<BR>\n" );
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document.write( "-2 = sqrt ( 4)\r<BR>\n" );
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document.write( "-2 = 2. This is obviously false, so x = -4 is an extraneous root. Let's check the other root, x = 3.\r<BR>\n" );
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document.write( "3+2 = sqrt(3(3) + 16).\r<BR>\n" );
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document.write( "5 = sqrt ( 9 + 15)\r<BR>\n" );
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document.write( "5 = sqrt ( 25 ) \r<BR>\n" );
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document.write( "5 = 5. That checks out so the solution is x = 3.  \r<BR>\n" );
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document.write( " </SPAN>\n" );
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