DERIVE for Windows version 5.05 DfW file saved on 30 Sep 2002
f(x):=- 5x^3 - 3x^2 + x + 4
coeff:=[4, 1, -3, -5]
hCross:=APPROX(41129032258064513/50000000000000000)
vCross:=APPROX(0)
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\viewkind4\uc1\pard\qc\cf1\b\f0\fs24 Finding the Roots of a Cubic Equation -
\par An Excursion into Mathematics History
\par \pard\b0
\par NOTE: \cf2 We recommend that before starting this lesson, you open a \f1 2D-\f0 Plot Window and then choose Window\f1 \f0 >Tile Vertically. In this way you can study the lesson and also see your plots \f1 side by side\f0 .
\par \cf1
\par \b Background:\b0
\par
\par Until the middle ages the problem of finding the exact solution (called a solution by radicals) for the roots of a cubic equation was not solved. It is believed that the mathematician Scipione del Ferro was the first to find a method for finding the solutions to such equations. Del Ferro did not publish his result, but communicated it to one of his students at the time of his death. The student, a fairly weak mathematician, hoping to gain fame challanged a fairly well known mathematician, Tartaglia (the name means "the stammerer". He had received violent sword wounds about the face and jaw from French soldiers when he was age 12.) with a set of 30 problems. Tartaglia countered with his own set of 30 problems. Tartaglia had a flash of inspiration after working on the problems for a while and solved them all in 2-1/2 hours, thus, winning the contest.
\par
\par At this point a third mathematician, Cardan, enters the picture. Cardan heard of Tartaglia's feat. He met with him and persuaded Tartaglia to divulge his secret. It took some convincing, but only after he swore never to divulge it. Cardan and one of his students worked on the result, improved it, and even found a brilliant method for solving quartic equations by radicals. Cardan also discovered that the method of solving cubic equations was known prior to Tartaglia's victory. He and his student visited del Ferro's son-in-law and saw the del Ferro's origional notebook containing the solution of cubics. Cardan felt that this released him from his vow. In his very famous book on mathematics, \i Ars Magna\i0 , he published both the formula for the cubic equation and quartic equation, giving credit to del Ferro, Tartaglia, and his student for their roles in the development of the formulas. A bitter feud ensued between Tartaglia and Cardan.
\par
\par \ul Exercise\ulnone
\par \cf3 1) Using a web search engine, enter the keywords: Cubic Roots, Solution, Cardan. You should find links to Tartaglia and del Ferro there also. Read about the history of the solution by radicals of cubic equations and write a brief report that expands on the bare facts given above.
\par \cf1
\par \b The Standard Form of a Cubic Equation:\b0
\par
\par First we begin with the form of a cubic equation as we know it.\f2
\par }
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\viewkind4\uc1\pard\cf1\f0\fs24 Arbitrarily choose values for the coefficients \i a\i0 , \i b\i0 ,\i c\i0 , and \i d\i0 or \f1 simplify\f0 the following \f1 expression\f0 to randomly generate their value. We recommend that you \f1 keep simplifying the expression \f0 until you get a value for the variable, \i coeff\i0 , that contains all non zero values.\cf2 \f2
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8xUser!coeff:=VECTOR(RANDOM(10)-5,i,1,4){\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Define a function, \i f\i0 (\i x\i0 ), as shown in expression #3, expand it so you can see the function as a cubic, and graph the result in a plot window. Adjust the plot scale so that you have a graph that includes both the maximum and minimum value of the function. Also, adjust it so that the graph fits nicely into the \f1 2D-\f0 plot window. This will be your example function throughout the remainder of our exploration.\cf2 \f2
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8User$f(x):=SUM(coeff SUB i*x^(i-1),i,1,4)"{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\b\f0\fs24 Shifting the \i y-\i0 axis\b0
\par
\par What do we mean by the phrase, "shifting the \i y\i0 -axis?" What we mean is that we replace \i x\i0 by a variable, say \i u\i0 , where \i u\i0 = \i x\i0 - \i s\i0 for some real number \i s\i0 . Then the line \i u\i0 = 0, or the soon to be new \i y\i0 -axis in a (\i u\i0 , \i y\i0 ) coordinate system, is a vetical line corresponding to \i x\i0 = \i s\i0 in the (\i x\i0 , \i y\i0 ) coordinate system. Observe this by writing the expression:\f1
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\viewkind4\uc1\pard\cf1\f0\fs24 Choose a value for \i s\i0 that is within the range of \i x\i0 -values of your graph and draw the graph. When you have seen the result, delete the vertical line. It will only clutter up the graph during later investigations.
\par \cf2
\par Why do we want to shift the \i x\i0 -axis? Our goal is to simplify the form of expression #1 by eliminating the squared term by making a substitution for \i x\i0 . To do this we first substitute \i u\i0 + \i s\i0 , the value for \i x \i0 in terms or \i u\i0 , \i \i0 into #1 and expand the result in tems of u.\f1
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\viewkind4\uc1\pard\cf1\f0\fs24
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8/GUsera*(u+s)^3+b*(u+s)^2+c*(u+s)+d=0S0kExpd(#5)~jt?;a*u^3+u^2*(3*a*s+b)+u*(3*a*s^2+2*b*s+c)+a*s^3+b*s^2+c*s+d=0w{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 What value for \i s\i0 will cause the coefficient of squared term to be zero? You can easily solve this on your own or you can highlight the subexpression of expression #6 that is the coefficient for \i u\i0 ^2 and click on the solve button. Now use the expression for your example polynomial that you graphed, substitute for \i x\i0 , solve for the value of \i s \i0 that will make the coefficient of the squared term 0, and substitute this value into your expression.
\par
\par Note that the resulting cubic in \i u\i0 has no squared term. What this means is that every cubic equation in the form of expression #1 can, by shifting the \i y\i0 -axis (substituting \i u\i0 +\i s\i0 for \i x\i0 ), be written as:\f1
\par }
8
Useru^3+m*u+n=0{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fcharset2 DfW5 Printer;}{\f3\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 after we divide both sides by the coefficient of \i u\i0 . We are making progress, but progress towards what? Before we continue. Let's explore a little more with an \f1 e\f0 xercise.
\par
\par \ul Exercise:\ulnone
\par \cf2 2)Draw the vertical line from expression #4 above substituting your value for \i s\i0 . Where does it cross the graph of your function? Note that this point is mid way between the local peak and valley in the graph. Verify this fa\f1 c\f0 t by moving the cursor and noting the \i x\i0 -coordinates of the peak and valley.
\par \cf1
\par Now we move on to the point where the big insight of del Ferro, and later Tartaglia, comes into play. We make yet another substitution. This time we substitute (\f2\'a9\f0 + \f2\'df\f0 ) for \i u\i0 in expression 7. When we highlight the cubed term expand the result we see expression #9. This looks like a mess! \f3
\par }
8User!(alpha+beta)^3+m*(alpha+beta)+n=0HUser C(alpha^3+(3*alpha^2*beta+3*alpha*beta^2)+beta^3)+m*(alpha+beta)+n=0>+{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fcharset2 DfW5 Printer;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 However, we can straighten it out. We need to find out what values for \f1\'a9\f0 and \f1\'df\f0 in terms of \i m\i0 and \i n\i0 . \cf2 Highlight the term in parentheses inside of the expression in \f1\'a9\f0 and \f1\'df\f0 and factor it in terms of \f1\'a9\f0 and \f1\'df\f0 .\f2
\par }
XJb Fctr(#9')y&1|?
=(alpha^3+3*alpha*beta*(alpha+beta)+beta^3)+m*(alpha+beta)+n=0n{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Now things start to make sense! If we set,\f1
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8Userm=-3*alpha*beta8Usern=-(alpha^3+beta^3)N{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fcharset2 DfW5 Printer;}{\f2\fmodern\fprq1\fcharset0 Times New Roman;}{\f3\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 we have two equations in two unknowns, \f1\'a9\f0 and \f1\'df\f0 . Furthermore, if we solve these equations, we can back track to the value a value for \i x, \i0 and by factoring obtain all values for \i x\i0 that satisfy expression #1. At this point we could use the solve button, but we want to see the brilliant method of del Ferro and Tartaglia.
\par
\par \ul Excercise\ulnone
\par \cf2 3) Using your transformed cubic equation, substitute the values for \i m \i0 and \i n\i0 into expressions #11 and #12. What are the resulting equations\f2 ?\f0
\par \cf1
\par From expression #11 we solve for \f1\'df\f3
\par }
8Z~User
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\viewkind4\uc1\pard\cf1\f0\fs24 Substitute this value into expression #12 and expand in terms of \f1\'a9\f0 . Multiply bolth sides of the resulting equation by 27\f1\'a9\super\f0 3\nosupersub and you will see that we have a quadratic in \f1\'a9\super\f0 3\nosupersub as shown below.\f2
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\viewkind4\uc1\pard\cf1\ul\f0\fs24 Exercises\ulnone
\par \cf2 4) Substitute \i t\i0 for \f1\'a9\super\f0 3\nosupersub in the above expression and find the roots of the resulting quadratic equation. The cube roots are the value for \f1\'a9\f0 . Expression #13 gives the value for \f1\'df\f0 . Recall that \i u\i0 = \f1\'a9\f0 + \f1\'df\f0 , and \i x\i0 = \i u \i0 + \i s\i0 . Thus, you have found at least one value for \i x. \i0 If you need to find the remaining two, you can use the division algorithm and solve the resulting quadratic. Note: the cubic may have two complex roots.
\par 5) Using your values for \i m\i0 and \i n\i0 , find the solutions for your personal equation. Check your answer first by tracing your graph to a root. Then check the complete result using the Derive solve command.
\par \cf1
\par So, even though Derive can solve cubic equations quite easily, we have taken a small excursion through history using some of the power of Derive in order to understand where the Derive solutions came from and to also appreciate the contributions of others to our mathematical heritage.
\par
\par \fs20 Carl Leinbach: leinbach@gettysburg.edu\f2\fs24
\par }