document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=559425>Question 559425</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need some help.  It simply says find the solutions but I am not sure what method it is asking me to solve it with.  I don't know if it means solve using the quadratic equation or if there is some other way I am suppose to be doing it.  Please help! Thanks.\r<BR>\n" );
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document.write( "Find the solutions:\r<BR>\n" );
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document.write( "4)	(m+3)^2 + 5(m+3) –14 = 0<BR>\n" );
document.write( "5)	(2x-1)^2 –4(2x-1) + 2 = 0\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #363407 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=363407\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> these are quadratic equations.<BR>\n" );
document.write( "instead of x, you have (m+3)<BR>\n" );
document.write( "instead of x, you have (2x-1)<BR>\n" );
document.write( "assign the expression inside the parentheses to a variable other than a variable that is being used in the parentheses and you'll see that more clearly.<BR>\n" );
document.write( "for your first problem:<BR>\n" );
document.write( "(m+3)^2 + 5(m+3) - 14 = 0<BR>\n" );
document.write( "let x = m+3<BR>\n" );
document.write( "your equation becomes:<BR>\n" );
document.write( "x^2 + 5x - 14 = 0<BR>\n" );
document.write( "this factors to:<BR>\n" );
document.write( "(x+7) * (x-2) = 0<BR>\n" );
document.write( "x = -7 or x = 2<BR>\n" );
document.write( "for your second problem:<BR>\n" );
document.write( "(2x-1)^2 - 4(2x-1) + 2 = 0<BR>\n" );
document.write( "let y = 2x-1<BR>\n" );
document.write( "your equation becomes:<BR>\n" );
document.write( "y^2 - 4y + 2 = 0<BR>\n" );
document.write( "this doesn't factor easily.<BR>\n" );
document.write( "use the quadratic formula on this.<BR>\n" );
document.write( "y = (-b +/- sqrt(b^2-4ac))/(2a).<BR>\n" );
document.write( "a = 1<BR>\n" );
document.write( "b = -4<BR>\n" );
document.write( "c = 2<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "-b = 4<BR>\n" );
document.write( "sqrt(b^-4ac) = sqrt(8)<BR>\n" );
document.write( "2a = 2<BR>\n" );
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document.write( "y = (4 + sqrt(8))/2 = 3.414213562<BR>\n" );
document.write( "y = (4 - sqrt(8))/2 = 0.585786438<BR>\n" );
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document.write( "since we set:<BR>\n" );
document.write( "y = 2x-1, then we get:<BR>\n" );
document.write( "2x-1 = 3.414213562 which results in:<BR>\n" );
document.write( "x = 2.207106781<BR>\n" );
document.write( "and we get:<BR>\n" );
document.write( "2x-1 = .585786438 which results in:<BR>\n" );
document.write( "x = 0.792893219<BR>\n" );
document.write( "there is no need to go back to the original equation of:<BR>\n" );
document.write( "(2x-1)^2 - 4(2x-1) + 2 = 0 because when we calculate 2x-1, we get y.<BR>\n" );
document.write( "since y is the value we used to solve the equation, we will get the same answer because:<BR>\n" );
document.write( "x = 2.207106781 times 2 minus 1 will equal 3.414213562 which is the value of y.<BR>\n" );
document.write( "similarly:<BR>\n" );
document.write( "x = 0.792893219 times 2 minus 1 will equal .585786438 which is the value of y.<BR>\n" );
document.write( "these 2 values of y are what we solved for originally.\r<BR>\n" );
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