document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1196803>Question 1196803</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve for k in the equation kx²-2kx+5=0, if the sum of the roots is 4. What is the product of the roots? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #829801 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=829801\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "kx^2-2kx+5 = 0 is in the form ax^2+bx+c = 0 with<BR>\n" );
document.write( "a = k<BR>\n" );
document.write( "b = -2k<BR>\n" );
document.write( "c = 5\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "According to Vieta's Formulas, we know that for quadratics the sum of the roots is -b/a and the product is c/a\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "sum of roots = -b/a<BR>\n" );
document.write( "product of roots = c/a\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The value of -b/a in this case is:<BR>\n" );
document.write( "-b/a = -(-2k)/k<BR>\n" );
document.write( "-b/a = 2<BR>\n" );
document.write( "So the sum of the roots is 2 and not 4. <BR>\n" );
document.write( "Your teacher made a typo somewhere.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "For instance, if k = 7, then we have <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=kx%5E2-2kx%2B5+=+0 ALIGN=MIDDLE  ALT=\"kx%5E2-2kx%2B5+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" > turn into <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=7x%5E2-14x%2B5+=+0 ALIGN=MIDDLE  ALT=\"7x%5E2-14x%2B5+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" > <BR>\n" );
document.write( "Turn to the quadratic formula to find the two roots being <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%287%2Bsqrt%2814%29%29%2F7 ALIGN=MIDDLE  ALT=\"%287%2Bsqrt%2814%29%29%2F7\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%287-sqrt%2814%29%29%2F7 ALIGN=MIDDLE  ALT=\"%287-sqrt%2814%29%29%2F7\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "The sum of these two roots is 2.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "---------------------------------------------------------\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Here's a proof of Vieta's Formulas when applying the quadratic case.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Let p and q be the two roots of a quadratic<BR>\n" );
document.write( "This means x = p and x = q are solutions to ax^2+bx+c = 0 aka x^2 + (b/a)x + (c/a) = 0<BR>\n" );
document.write( "We can get everything to one side to get x-p = 0 and x-q = 0<BR>\n" );
document.write( "Then apply the zero product property getting (x-p)(x-q) = 0\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Then expand and follow these steps<BR>\n" );
document.write( "(x-p)(x-q) = 0<BR>\n" );
document.write( "x(x-q) - p(x-q) = 0<BR>\n" );
document.write( "x^2-qx - px + pq = 0<BR>\n" );
document.write( "x^2 - (p+q)x + pq = 0\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Comparing that to x^2 + (b/a)x + (c/a) = 0 shows:<BR>\n" );
document.write( "b/a = x coefficient = -(p+q) = sum of the roots<BR>\n" );
document.write( "c/a = constant = pq = product of the roots\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Then as one last final step we turn the <BR>\n" );
document.write( "b/a = -(p+q)<BR>\n" );
document.write( "into<BR>\n" );
document.write( "p+q = -b/a<BR>\n" );
document.write( "to help match the original statement made about Vieta's formulas<BR>\n" );
document.write( "</font> </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );