Creating Rational Equations that have R ational Solutions. By Alan Roebuck Chaffey College

Transcription

1 Creating Rational Equations that have R ational Solutions aversion Î /' y Alan Roeuck Chaffey College alanroeuck@chaffeyedu th Page )'! of Elementary and Intermediate Algera & edition y Tussy and Gustafson contains a word prolem leading to an equation of the form + The prolem is this: Worker 1 can complete a jo in hours working alone Worker 2 working alone can complete the same jo in + hours and it takes hours to do the same jo when they work together How long does it take Worker 1 to finish the jo working alone? When the equation is cleared of fractions we get a+ +! How can the instructor choose the parameters + and so that the equation is factorale that is the solutions are rational? This is not an isolated question Many rational equations when cleared of fractions yield quadratic equations It's useful for these quadratics to have rational solutions ecause the student may not yet know the quadratic formula or completing the square And the equation may have arisen from a word prolem in which case we prefer a solution that doesn't require a calculator How can we choose the coefficients so that the rational equations will have only rational solutions? elow we examine several common types of rational equation which are quadratic when cleared of fractions We show that there are clear criteria for the values of the coefficients of the equations that will make the solutions rational With the needs of the math instructor in mind we also show rules (many of them quick and easy) for determining these rational-solution-making coefficients In fact we will learn that you can start with any factorale quadratic equation and then generate a rational equation whose reduced equation is the quadratic you started with You can even specify that some or all of these solutions are extraneous These results may also e of value for designers of mathematical educational software: Algorithmically-generated equations must e feasile for the student to solve and this will often require rational solutions Preliminaries Throughout this paper all equation coefficients will e assumed rational Some terminology: Definitions a) Given a rational equation the reduced equation is the equation cleared of fractions Page ) A quadratic-type rational equation has a quadratic reduced equation )'! of Tussy and Gustafson contains two types of rate prolems The first type was mentioned aove: p 1 Copyright!' y Alan Roeuck May not e reproduced without permission

2 Part I) A rate-of-work word prolem The Prolem When Workers 1 and work together a jo takes hours When Worker does the same jo alone it takes + fewer hours than if Worker does the jo alone How long will it take Worker to do the jo himself? We can solve the prolem y considering the rate of work for three scenarios: The workers work together Worker 1 works alone and Worker 2 works alone: jo Net work rate Vnet hours Worker 1's rate V Worker 's rate V + When two or more workers work together their net work rate equals the sum of their individual work rates: V V Vnet that is + Reduced equation: a+ +! Solutions: + È + % In the typical textook rate prolem + and are integers More generally we require them only to e rational Typically + and are positive (and the rate prolem will have no valid solution if is negative) ut we will not require this The equation is also a valid rate prolem if + is negative (meaning that Worker 2 is slower than Worker 1 so that + is replaced y l+lñ and it can e used in the algera class as a non-word-prolem rational equation even if is negative Since + and are rational solutions are rational iff the discriminant is the square of a rational numer And H + % + ÐÑ Oviously a sufficient condition for rational solutions is that k+ k and k k are the legs of a Pythagorean triple such as aß % or a&ß % ut this is not a necessary condition We could have for example + ( and ( That is a+ß ( aß % ß and then + ÐÑ Š Š % * ' & & ( ( %* %* %* Š ( arational We could even have a+ ß & a& ß Š ß & and then!%%* % + a Š & & Š & arational These examples show that a more general sufficient condition for rational solutions is that ak+ k k k 7 8 times the first two components of a Pythagorean triple where 7 and 8 are nonzero natural numers ut is this a necessary condition? Since coefficients must e rational yes To prove it let's first define some terminology: Definition A rational (respectively integer or natural) Pythagorean triple is any triple of nonzero rational numers (respectively integers or natural numers) a+ß ß - for which + - In other words + is the square of a rational numer p 2 Copyright!' y Alan Roeuck May not e reproduced without permission

3 So it makes sense here to define an (unqualified) Pythagorean triple to e any triple of nonzero rational numers a+ß ß - for which + - It is worth emphasizing that in a Pythagorean triple as defined here + or - could e negative or fractional The next definition is the most important one: Definition A Pythagorean pair a+ß is the first two components of a Pythagorean triple So in a Pythagorean Pair + and are rational numers for which + arational numer What must + and equal to meet this requirement? - / Since they are rational and nonzero + and 0 where -ßß/ßand 0are nonzero integers Therefore + ˆ - / Š 0 a-0 a/ a0 This is rational iff a-0 a/ ainteger that is ak-0kß k/ k are the first two components of a natural Pythagorean Triple Therefore + is the square of a rational numer iff ak-0kß k/ k 8 aß C where and C are the first two components of a primitive Pythagorean Triple Divide y k0k: 8 k0k ak -0 k ß k / k k0k a ß C Š - / ß¹ 0¹ 8 aß C k0k ak+ß k k k 8 aßc k0k 7 That is a+ß 8 a ß Cß where 7 8Á! are natural numers This is the necessary and sufficient condition for + to e the square of a rational numer 7 We just proved that every Pythagorean Pair equals 8 a ß C where 7 and 8 are natural numers and aß Care the first two components of a primitive Pythagorean Triple We might call this the Pythagorean Pair Theorem: Given nonzero rational numers + and a necessary and sufficient condition that + e the 7 square of a rational numer is that a+ß 8 a ß C where 7 8are nonzero natural numers and aß Care the first two components of a primitive Pythagorean Triple So from now on A Pythagorean Pair is any ordered pair a! ß such that 7 a! ß 8 a ß C where aßc are the first two components of a primitive Pythagorean Triple and 7 and 8 are nonzero natural numers p 3 Copyright!' y Alan Roeuck May not e reproduced without permission

4 Note also that A nonzero rational numer times a Pythagorean Pair equals another Pythagorean Pair since nonzero natural numer 5 8 a ß C 58 a ß C another nonzero natural a ß C Pythagorean Pair Therefore we can characterize all equations having the form found in word prolem Type I which have rational solutions: Theorem I Rational solutions of the work-rate equation The equation + has rational solutions iff a+ß is a Pythagorean Pair Note that the theorem remains valid for equations of the form - + : Let?-À??a-+ Rational solutions iff a-+ß is a Pythagorean pair Also the theorem remains valid for + ecause if a+ß is a Pythagorean Pair so is a +ß Let's oserve how the solutions of the equation depend on the parameters + and : Example Take the Pythagorean Pair a! ß a&ß Then + & and so that ' Equation: & ' Reduced equation: (!! Solutions: &ß Every natural Pythagorean pair contain one odd and one even numer so can always e a whole numer even if we restrict ourselves to integer Pythagorean Pairs Reversing the signs of the constants: Reverse oth signs: Equation is & ' Solutions: &ß : Opposites of the previous solutions Reverse just one sign: Equation is & ' Solutions:!ß Reverse just the other sign: Equation is & ' Solutions:!ß Compared with the previous equation we see that again reversing the signs of oth constants reverses the signs of oth solutions % % Example Start with the Pythagorean Pair a! ß Š ( ß ( Then + ( and ( so that ( The equation is: ( ( Reduced equation: %* ('! Solutions: ( ß ( Here the solutions are + and ut this does not usually occur ( % Example Starting with the Pythagorean pair a(ß % let + and ) so that % Equation: % ( % Reduced: ( )! Solutions: (ß p 4 Copyright!' y Alan Roeuck May not e reproduced without permission

5 Terminology: the rationalizer Consider the equation + In this paper we're not trying to solve the equation Instead we're trying to rationalize the solutions So it is natural to make the following Definitions Given the equation + ß an ordered pair a+ß for which solutions for are rational numers is called a rationalizer of solutions of the equation or more riefly a rationalizer An equation with rational solutions is a rationalized equation Strictly speaking we are rationalizing the solutions of the equation ut since we work directly with the equation which defines the solutions only indirectly it is natural to think of rationalizing the equation rather than the solutions Therefore we can say that a+ß is a rationalizer of More generally we have the following + iff a+ß is a Pythagorean Pair Definition Let JÐß+ßß-ßáÑ! e a rational equation with as the independent variale and +ßß-áas parameters (ie coefficients) If! ß ß ßá are values of respectively +ßß-á then a! ß ß ßâis a rationalizer of the equation iff the solutions of JÐß! ß ß ßáÑ! for are rational numers The suject of this paper is therefore the rationalizing of certain classes of rational equations We could therefore reword Theorem I to say that if rationalizes the equation + a! ß is any Pythagorean Pair then a+ß Š! ß Or:! has rational solutions for iff a! ß is a Pythagorean Pair Rationalized equations are plentiful Creating equations with rational solutions from Pythagorean Pairs is less restrictive than you might first think: Theorem Given any nonzero rational numer - there is a Pythagorean Pair whose first component is - and a Pythagorean Pair whose second component is - Proof Let aß C e the first two components of a primitive Pythagorean Triple and rememer that any nonzero rational numer times a Pythagorean Pair is another Pythagorean Pair Therefore - aß C is a Pythagorean Pair with first component - - C aß C is a Pythagorean Pair with second component - Corollary Since there are infinitely many primitive Pythagorean Triples in the equation + given any rational numer there are an infinite numer of choices of + that will make solutions rational And given any rational numer + there are an infinite numer of choices of that will make solutions rational We could say that the set of all rationalizers a! ß of the equation + is dense in p 5 Copyright!' y Alan Roeuck May not e reproduced without permission

6 In the word prolem recall that + is how much more time or less time one worker takes compared to the other If + is a fraction the expression + will cause difficulty for the student Therefore in the word prolem we prefer that + e an integer ut can e a fraction without causing major difficulties for is easy Therefore we have the following result: Theorem II Creating a good work-rate prolem with + specified Let aßc e the first two components of a primitive Pythagorean Triple and let + e an integer To rationalize the rate equation + choose the Pythagorean Pair a +ß + a +C ß C ˆ +ß Therefore +C so that the equation is + +C Since there are infinitely many primitive Pythagorean Triples there are an infinite numer of ways to rationalize the equation for a given + + ac + È C Solutions of this equation are Since aßcis a Pythagorean pair È C is rational so solutions are rational numers Example Suppose +) Using aßc aß% we get +C ) % ' Equation: ) ) ' Solutions: 'ß (The solutions and the equation parameters are similar and there is proaly an underlying reason) This is the answer to the following word prolem: When Joe works alone it takes him 8 fewer hours to do a jo than when o works on the same jo alone Working together it takes them & hours to do the same jo How long does it take when o works alone? Example This time specify the time it takes when they work together Suppose we want For rational solutions of the equation we require a+ß Š +ß to e a Pythagorean Pair Start with aß % and multiply y 6 to get Š 2 ß Therefore + and Equation: that is 11 Solutions: ß ' In the rate prolem we can specify (the time required to complete the jo when the workers work together) to e any rational numer and then we can find infinitely many rational + -values for which rational solutions exist Example Let + We need a Pythagorean Pair of the form a ß : &' & a&ß Š ß & is the Pair we need So let a+ß Š ß &' () & Then & and the equation + is: & Reduced: Solutions: ' () &!%! *ß & If we start with aß % and let + & we get a more manageale equation: p 6 Copyright!' y Alan Roeuck May not e reproduced without permission

7 & ß % &ß! +ß! a Š a so Equation: Reduced: Solutions:! &! & &!! &ß Part II) A Uniform-motion word prolem This is the second type of word prolem on page )'! of Elementary and Intermediate Algera & edition y Tussy and Gustafson Examples include questions )* and *! The prolem An oject travels a distance of miles at an unknown speed It could have traveled the same distance in 2 fewer hours if the speed had een increased y + miles per hour Find the first speed So first speed atime required when traveling at speed atime required when traveling at speed + + Equation: Divide y : 2 Define and the equation is: + + È + %+ Reduced equation: ++! Solution: As efore + and are rational so that solutions will e rational iff + %+ is a square Typically + and are positive ut as in Type I this is not necessary And the prolem may not have a real solution if + and have opposite signs If %+ is positive then it is the square of È %+ and + %+ + Š È+ Therefore Theorem III Rationalizing the uniform motion word prolem Given + positive a sufficient condition for rational solutions of + is that Š +ß È + e a Pythagorean pair We can refer to the components of a Pythagorean Pair as legs: Aside from sign first choose + to e one leg and then choose to satisfy È+ other leg And rememer that + can e any rational numer Example Let + and È+% Then È%ß so È % and : Equation: % Reduced equation: *! Solutions: %ß When the sign of one constant is reversed we get complex solutions À 3 È ( % Reduced equation: *! Solutions: 3 È ( % Reduced equation: *! Solutions: When the signs of oth constants are reversed we get: th p 7 Copyright!' y Alan Roeuck May not e reproduced without permission

8 % Reduced equation: *! Solutions: %ß These are the opposites of the solutions when oth constants are positive Example Start with the Pythagorean pair a&ß and take oth + and to e negative for then %+ will e positive Divide y and make + negative: + & and È + % É & & Then kk 4 so kk %ßso & & % Equation: & Reduced: & a &! Solutions: ß In general let a e a Pythagorean Pair Then we require that and È! ß +! + that is %+ %! In other words a+ß Š! ß %! rationalizes + : Theorem IV Explicitly rationalizing the uniform-motion word prolem 4! If a! ß is a Pythagorean Pair then! has rational solutions The solutions are! É! Non-Pythagorean methods part I: the method of equal discriminants We are not restricted to Pythagorean Pairs when seeking to rationalize an equation Other approaches are possile Example Start with the equation ( & a a &! Solutions are &ß The discriminant is H a ( % a & %* %! '* Recall that the equation + reduces to a quadratic equation whose discriminant is H + %+ We could refer to + %+ as the discriminant of the rational equation: Definition The discriminant of a rational equation is the discriminant of its reduced equation if the reduced equation is quadratic As shown elow can create an equation of the form + whose reduced equation also has H '* Therefore its solutions are rational For the equation ( &! we have H %* %! The Method of Equal Discriminants is to set this discriminant equal to the discriminant of our rational equation and solve for the coefficients: + %+ %* %!! Comparing terms we get + %* so that + (ß and +! so that ( ( If +(ßthis gives the equation (! Solutions are and! which are times the solutions of the equation with which we started Š & and And that equation has a leading coefficient of p 8 Copyright!' y Alan Roeuck May not e reproduced without permission

9 ( If + (ßthis gives the equation (! Solutions are and! These are times the original solutions % Example Start with a % a ' % %! Solutions are 'ß ( ' Therefore + % and +(ß so % ( H% % (+ %+ ( If we choose the rational equation is % ' Solutions %ß ) % These are times the solutions of the original equation Š ' and and that equation has a leading coefficient of ( Choosing reverses the sings of the coefficients: % ' : Solutions: )ß % This suggests that starting with a leading coefficient of one will give our rational equation the same solutions as the quadratic with which we start: Example Take the previous starting quadratic and divide y three: % % % Š a' )! Same solutions: 'ß Use this discriminant to rationalize our rational equation: % HŠ % ) Therefore + and +) that is + ( % ) + %+ ( If we choose the equation + is % % Reduced: % % Solutions: ß ' The opposites of the original solutions ( % If we choose the equation is Solutions: ß ' Same as original solutions % Example Start with!a a& (! H( % a! + %+ Therefore + ( and +! so! ( ( Choosing the sign the equation is (! Solutions ß & Same as original solutions Let's prove that this always happens And we don't need the method of equal discriminants to prove it We can use an easier method: p 9 Copyright!' y Alan Roeuck May not e reproduced without permission

10 Non-Pythagorean methods part II: the method of equal coefficients Choose rational numers - and and form the simplest quadratic equation having these as solutions: The starting quadratic:!a-a a- - Recall that for the rational equation + the reduced equation is ++! Compare coefficients: a- -! + +! We must choose + and so that the rational equation will have the same solutions as the starting quadratic The simplest way to do this is to require that the two quadratics have the same coefficients This means a- and + - that is + - The rational equation is According to my TI-89 the reduced equation is a-a! Solutions are - and the original rational numers we chose Oserve that this method fails if the leading coefficient is not one: Ea< a< E E a< < E< <! Compare: + +! Won't work unless E ut the method of equal discriminants continues to work if the leading coefficient is not one Given any quadratic equation with rational roots we can create a rational equation of the form + whose roots are multiples of the roots of the quadratic Theorem V The Method of Equal Discriminants Let E FG e any factorale trinomial with discriminant F %EG If + and are chosen so that EG + %+ F %EG that is + F and F then + has rational solutions - In particular if - and are any rational numers - - has - and as its solutions This method is useful in that it allows us to specify oth solutions in advance In terms of the word prolem which generated this equation we can specify in advance the first speed: Example Recall that the equation is + where + is how many miles per hour faster the second trip is compared to the first And 2 where is the distance of oth trips and 2is how many fewer hours the second - & trip takes Suppose we want the first speed to e & mph Then - & And - & so if ß the equation is ( (! ut this is not valid for the word prolem ecause solutions are and & so the second speed ( would then e negative p 10 Copyright!' y Alan Roeuck May not e reproduced without permission

11 - &) Try again: -& and )Then - and - %! %! The equation is %! and the solutions are & and ) This equation results from the following word prolem: An oject travels %! miles at an unknown speed It could have traveled the same distance in fewer hours if the speed had een increased y miles per hour Find the first speed Why does the method of equal discriminants work? egin with the easy case: leading coefficient of one Let < and < e any two rational numers and start with this quadratic equation: The Starting Quadratic a< a< a< < < <! Its discriminant is H a< < %< < U Recall our rational equation: + Reduced: V + +! Discriminant: H + %+ Solution of the rational equation: W V + È + %+ In the method of equal discriminants we set the discriminants equal to one another and solve for + and : H H that is + %+ a< < %< < V U << << Therefore + a< < and + << that is + < < Let's calculate the solutions of the rational equation for these values of + and : Lemma For any quadratic equation E F G! the discriminant is HE a< < E a< < That is the discriminant is the product of the square of the leading coefficient and the square of the difference of the roots Proof If < and < are the roots of the equation then E FG Ea< a< Ea a< < < < E E a< < E< < This equation has discriminant HE ˆ a< < %< < E a< < QED Returning to our rational equation the solutions are + È+ %+ + ÈHV + ÈHU W ecause H H V V U Since our starting quadratic equation had a leading coefficient of H a< < And + a< < So if + < <ßW V U a<< É a< < a << k < < k a<< k< < k a<< k< < k WV and p 11 Copyright!' y Alan Roeuck May not e reproduced without permission

12 a<< < < a<< < < If < <!ß then solutions and < and < a<< a< < a<< a< < If < <!ßthen solutions and < and < In either case if + < <ß solutions of the rational equation are the opposites of the solutions of the original quadratic And note: Since the original quadratic equation is a< < < <! +< < means that +opposite of coefficient of And if + a< < the coefficient of then W V << k<< k << k<< k and << << << << If < <!ß then solutions and a < < and << a<< << << If < <!ß then solutions and < and < In either case if + a< < coefficient of ßthen solutions of the rational equation are the same as the solutions of the original quadratic In what way would the analysis change if we started with a quadratic whose leading coefficient is not one? Let < and < e two rational numers and start with the quadratic equation: Its discriminant is: HU E a< < E< a a< E a a< < <<! Set this equal to the discriminant of the rational equation: Therefore + E a< < and + E < < + %+ Eˆ a< < %< < So + E< a < and we do not need an expression for For the analysis elow we will need the fact that E a< < ke ka< < + The solutions of the rational equation are : + ÈH + ÉE U a< < + ke kk< < k WV ke ka< < ke kk< < k a< < k< < k If + ke ka< < ß WV kek We saw aove that the racketed expression equals < and < Therefore If +ke ka< < ßsolutions of the rational equation are ke k times the opposites of solutions of the original quadratic ke ka< < ke kk< < k a< < k< < k And if + ke ka< < ß WV kek p 12 Copyright!' y Alan Roeuck May not e reproduced without permission

13 We saw aove that the racketed expression equals < and < Therefore the solutions of the rational equation are + ke ka< < kek We can simplify our results y making the assumption that times the solutions of the starting quadratic equation if E! In that case we have these results: EG If + E< a < the coefficient of and + then solutions of the rational equation are < and < Theorem VI The Method of Equal Discriminants Updated Let E F G e any factorale trinomial with E! EG EG If + F and + F then the solutions of F + that is F EG are E times the solutions of E FG! Example Start with a& a ( & %! Solutions (ß & F F For the rational equation F EG let F and EG & a% (! Equation: (! Reduced: ( a & a! Solutions: &ß These are & ( and & Š & which are & times the solutions of the original quadratic Methods of equal discriminants and equal coefficients for the first type of equation Could we make similar calculations on our first equation type +? Recall that its reduced equation is a+ +! Using the method of equal coefficients let's try to make the reduced equation equal to a- a! so that its solutions will e - and : a- -! a+ +! Taking the rational numers - and as given we attempt to find rational + and so that the aove quadratic equations are identical If not for the factor of the solutions for + and would e trivial ut after much unsuccessful calculation the author conjectures that there exists no such rational solutions for + and + So let's just try for rational solutions of discriminant of the rational equation is + rather than specifying solutions in advance Recall that the H+ % The rational equation will e rationalized if + and fulfill the condition that + % F % aeg where E F G is factorale This would require p 13 Copyright!' y Alan Roeuck May not e reproduced without permission

14 + F and È EG ecause we require rational coefficients EG must e a square Example Start with a a '! Here EG ' The quickest way to make EG a square would e to multiply it y This can e effected y going ack to the factored equation and multiplying the -terms y and respectively: a a ' & ' Now EG ' And the solutions are So we require that F %EG & %% + % Compare terms: + & so that + &ß and ' so that ' Choose the rationalizer a+ß a&ß' : + ecomes & ' Solutions:!ß This method requires starting with an equation of the form a- a - so that EG - Without loss of generality we consider the equation a- a- - a - -! H a - a- This equals a - ut we don't need this fact Set the discriminant of + equal to a - a- : + a a - a- Therefore + a - and - If we choose oth signs positive the equation is - - According to my TI-89 solutions are - a- and a- These are rational That is Theorem VII For any rational numers - and the equation - - has rational solutions The solutions are - a- and a- p 14 Copyright!' y Alan Roeuck May not e reproduced without permission

15 Specifying the solutions at the eginning Given two rational numers < and < ß we can create an equation of the form - - whose solutions are < and < if we can solve the system < - a - and < a- for - and in terms of < and < This is the nonlinear system - - < The formal solution for a-ß is quite complicated: - < So that the expressions will fit the page without eing too small let < and C<Then: a-ß Œ ˆ È 'CC CC ÉÈ 'CC C ß ÉÈ %C 'CC C or a-ß Œ ˆ È 'CC CC ÉÈ 'CC C ß ÉÈ %C 'CC C Not only are the expressions cumersome ut we must have - and - rational It is not ovious how to make this so In fact it appears to e impossile to give expressions for rational - and that guarantee that the solutions of the rational equation are < ß< ut we can produce a manageale rule if we drop the requirement that oth solutions e specified in advance If we only require that < e a solution and we don't care what the other solution is as long as it's rational we only need - and to meet the requirement < - a- that is -- < Then - can e any rational numer and <- - Recall that we are working with the rational equation - - The equation contains the expressions <- < - - Š - - < - and the expression - < - oth of which are rational if < and - are rational Therefore the rational equation has rational coefficients The equation is therefore - - < - We have proved: < - < Theorem VIII For any nonzero rational numer - < is a solution of < - < < - < - < - According to my TI-89 the solutions are < and - - % - < - < - ŒOserve that if we can solve the equation < - for - then the second solution will e < % p 15 Copyright!' y Alan Roeuck May not e reproduced without permission

16 < Example Start with < % and - Then - < % % )) - * and < - % Equation: )' %ß )) Solutions: * * To make the equation manageale we don't want < - < - to e too complicated So it would e etter to choose < negative < Example Let < and - Then - < a - & and < - ' Equation: & ' Solutions; ß! A second solution-in-advance method The author accidentally otained the following alternate expression for an equation of this form with one solution specified This expression has no free parameter - For the equation - - he accidentally wrote down the solution as -- a and -- a The first solution is correct ut the second should e - a ut if we try to find values of - and which make < - a - and < -a- we get a useful expression This will create an equation with as one solution < Expand: Therefore < - - and < - - < < - and < < - The equation we are trying to create will contain - and - and we need oth to e rational: Therefore and - << << and - so that a<< a<< %- a< < << a<< << - a< < < < - << oth are rational And these equations are valid even if - - < < Therefore since the equation is Theorem IX If < and < are rational solutions of < < are rational And one solution is < < < < < < < In practice we should proaly pick a<ß< so that << is an integer This will e true if < < p 16 Copyright!' y Alan Roeuck May not e reproduced without permission

17 Example Suppose < * and < ( Then < < * ( 2 << ' and << ) The equation is is < < < < ' Solutions are ) *ß &' < < Only our < is a solution Our < is not actually a solution Suppose we exchange the roots: < ( < * The equation is now Again our < is a solution ut not our < ' ) Solutions are ( ( Example Let < &ß < ) Then < < < < << )! and The equation is Once again only our < is a solution!% )! Solutions: &ß According to my TI-89 the solutions of < a< < < < < < are < and < < < < Summary We were trying to find values of + and so that the equation + will have the same discriminant as the equation a- a- - a - -! therey guaranteeing rational solutions We found that this would e so if and only if + a - and - There are a total of four possiilities for our rational equation: c a - d - We explored the first possiility - - ß ut there are three others: For - - This comes from replacing - with - in the previous equation For - - This comes from exchanging - and in the original equation For - - This comes from replacing - with - in the previous equation p 17 Copyright!' y Alan Roeuck May not e reproduced without permission

18 Non-Pythagorean methods part III: the method of free parameters We continue to discuss the equation + There is a simpler approach to creating an equation with rational solutions: Require its discriminant H+ %+ to e the square of a rational numer If we choose any rational numer -(the free parameter) and require that + %+ - then + %+ will e the square of a rational numer and the rational equation will have rational solutions This equation is linear in so we can solve for : -+ %+ Given any rational numers - and + calculate in the aove way and the equation is rationalized - is a free parameter ecause it does not appear in the equation It can e any rational numer except that cannot e zero (ecause of the term in the equation) and therefore - cannot e + Therefore we have this result: Theorem X Method of free parameters for the uniform-motion word prolem Let +Á! and -e aritrary rational numers except that - Á + If -+ then %+ + has rational solutions Although it is not necessary we prefer that - + e a multiple of % so that is an integer A sufficient condition for this that + and - oth e even - + ' Example Choose + ß - ' Then %+ % ) % The equation is % Reduced: )! Solutions: %ß - + % & * *! Example Let + & and -% Then %+!!! Equation: & * * Reduced:!!! %&! that is & a%! *! Solutions: ß To summarize our methods for equation Type II: Theorem XI Summary of methods to rationalize the uniform-motion word prolem If + and are rational the equation has rational solutions if: + I) II) Š +ß È+ is a Pythagorean pair or EG + F and F where E FG is factorale and EG is negative or -+ II) + and - are any rational numers aexcept that + Á! and + Á - and %+ IV) If - and are rational they are the solutions of p 18 Copyright!' y Alan Roeuck May not e reproduced without permission

19 Here's a typical rational equation from Algera I: Part III) A rational equation with one quadratic denominator &' Note that the denominator on the right is a a so that the LCD will not e cuic or quartic The solutions are and ' ut simply changing the second numerator from to changes the solutions to ' È This raises the question How can we ensure rational solutions for this type of equation? At first sight the numer of possile coefficients might appear to e too great to admit of any useful rule ut such is not the case Ensuring rational solutions is relatively easy and the multiplicity of coefficients gives us more control over the appearance of the equation We egin our analysis with the simplest version of the aove equation: + a + Reduced equation: +! If + is rational then solutions are rational which is a useful (aleit mundane) result for the teacher: Theorem XII If + is rational the equation + a+ has the rational solution + ut we were looking for a quadratic another coefficient reduced equation so we have oversimplified the equation type We need We won't consider + a + a for this is just a translation of the equation immediately aove: let Cß and we get C C+ a CC+ a a which has the same form Similarly we don't consider + a + which also reduces to a linear equation No to get a quadratic equation we need a factor of in the second or third numerator For example: Equation: È % a+ + a+ Reduced: +! Solutions: Since + and are rational we need % a+ arational in order for solutions to e rational We could use the method of free parameters: Choose to e any rational numer choose any other rational numer - and then choose + to meet the requirement that % a + - %- that is + % We won't give any examples yet ecause we're just illustrating the general technique We're working our way up to a general formula that will encompass all of these preliminary cases What if we include another coefficient? The equation - + a + reduces to: a- +! Solutions: - Éa- % a+ p 19 Copyright!' y Alan Roeuck May not e reproduced without permission

20 We need a- % a+ arational One method would e to choose +ß and - so that ˆ -ß È a+ is a Pythagorean triple Since there are three unknowns ut only two equations infinitely many solutions exist Or we could use the method of free parameters: First choose any rational numers - and is the free parameter ecause it is not in the equation Then choose + to meet the requirement that a- % a+ a- % that is + % We could also use the method of equal determinants: Let determinant HF %EG and set E F G a- % a + F %EG e a factorale trinomial with Therefore - F and a+ EG Choose +ß and -to satisfy these equations How aout more coefficients? The equation - / + a + has the solutions - Éa- % a/+ Either require ˆ È -ß a/+ to e a Pythagorean triple or require a- % a/+ 0 for some rational free parameter 0 Since this last equation contains just a single instance each of + and / it is easiest to choose all coefficients other than + or / to e rational choose any rational 0 and then calculate the remaining coefficient from this equation Or use the method of equal determinants ut let's look at the general case: One quadratic denominator: the general case Let's relael our coefficients to a more natural system: Consider the equation - / a + a Note the laeling scheme: + and are the opposites of the zeros of the denominators and - through 2 are the coefficients of the (linear) numerators from left to right Note also that the LCD is a+ a The three denominators contain only two distinct linear factors Reduced equation: a-/ a+/-01+02! Discriminant: H a+/-01 % a-/ a+02 Reduced equation could e linear One way to ensure rational solutions is for the reduced equation to e linear This requires -/! that is / - This means that the two numerators on the left side of the rational equation have opposite - coefficients: - and -0 Theorem XIII Linear reduced equation for one quadratic denominator - /0 12 The equation + a + a reduces to linear iff -/! that is the coefficients of in the left-side numerators sum to zero p 20 Copyright!' y Alan Roeuck May not e reproduced without permission

21 Example & %( & a a& Reduced: & %! It's easy to construct a formidale-looking rational equation that reduces to linear What if -/Á!? One option would e to see the discriminant as eing H a+/-01 c% a-/ a+02d a+/-01 % a-/ a2+0 a+/-01 Èa-/ a2+0 if the radicand is positive Therefore we need ˆ +/-01ß Èa-/ a2+0 to e a Pythagorean triple Since there are eight coefficients and only two constraints infinitely many solutions exist for every given Pythagorean triple ut the method of free parameters is easier: Note that the expression for H only contains a single (and linear) instance of 2 (In the original rational equation 2 is the constant term in the numerator on the right) Therefore we can choose another rational numer call it 3 set H 3 and then solve for 2 This means: a+/-01 % a-/ a+02 3 and therefore a+/-01 3 % a-/ a+02 that is a+/-01 3 %-/ a +02 and so 2+0 a +/-01 3 %-/ a This leads to: Theorem XIV: Method of free parameters for one quadratic denominator Consider the rational equation - / a + a If /-! the reduced equation is linear and there is one rational solution If /-Á! we must find eight coefficients To make solutions rational choose the seven coefficients + ß-ßß/ß0ßand 1to e any rational numers and then choose 2+0 a+/-01 3 %-/ a 3any rational numer & 2 Example Let's start with a a In other words we aritrarily specify the coefficients + through 1 What must 2 e for solutions to e rational? Comparing with the formula we see that + ßß-ß!ß/!ß0&ßand 1 Therefore we require a!!& 2 &! 3 ' %! a & 3 '3 ) & ) 3 can e any rational numer ut we prefer to make 2 an integer If 3 ß then 2 & ) * and the & * equation is a a Reduced: ' %! Solutions: ß ut is extraneous so p 21 Copyright!' y Alan Roeuck May not e reproduced without permission

22 We chose all of the other coefficients at will and then only ensuring that 2 is an integer 2 was constrained The only difficulty might e We can also apply the method of equal discriminants If the discriminant of a factorale quadratic is HF %EG then set this equal to the discriminant of our rational equation: And so a+/-01 % a-/ a+02 F %EG +/-01 F and a-/ a+02 EG Choose aritrary rational values for all coefficients except two and determine the two remaining ones y the aove equations The author recommends that 1 and 2 e the last two Creating extraneous solutions The equation in the last example had one extraneous solution ut that was just fortuitous (Or unfortunate depending on your point of view) We did not set out to create an extraneous solution ut the algera instructor sometimes needs to create rational equations with extraneous solutions and the aove analysis can e extended to allow this to e done delierately More generally we will show how to pick two rational numers and then construct a rational equation having those numers as solutions And either one or oth can e extraneous! egin y choosing a factorale quadratic equation: E FG! This factors to Ea< a<!ß where < and < are the solutions We will construct a rational equation of the form found in Theorem XIV whose reduced equation is E F G! Therefore the solutions of the rational equation will e < and < As we saw the equation - / a + a has reduced equation: a-/ a+/-01+02! We want this equation to e the equation we chose: E F G! If so the solutions of the rational equation will e < and < Therefore we must choose the coefficients + through 2 to satisfy three conditions: -/E +/ F +0 2 G Since there are only three equations determining eight variales there are infinitely many solutions Let's solve them in an orderly way: egin y choosing the denominator coefficients + and If + < then one denominator factor will e < and < will e an extraneous solution If desired we can also choose < and then oth solutions of the reduced equation will e extraneous If only < is to e extraneous and + < then we can choose four of the other coefficients aritrarily and then the three equations determine the three remaining coefficients For example we could assign values to ß ß / and 1 and then calculate: -E/ 0F1+/- 2+0G e sure to solve them from top to ottom p 22 Copyright!' y Alan Roeuck May not e reproduced without permission

23 Oserve that if we choose integer coefficients (including Eß F and G) the remaining coefficients will also e integral Example Choose the reduced equation to e a a &! Therefore aeß Fß G aß &ß And the solutions are and We will create a rational equation of the form in Theorem XIV whose solutions are and Let's set + Since one denominator factor will e ß this makes an extraneous solution Next choose Since -E/ let's set /! so that -! The remaining coefficients are ß 0ß 1 and 2 0 and 2 are determined y the aove equations so we can choose and 1 aritrarily Let's choose aß1 a&ß Then 0 F1+/- &a! & and 2+0G a a& Altogether a+ß ß -ß ß /ß 0ß 1ß 2 aßßß &ß!ß ßß The equation is thus & that is & a a ' Multiplying y the LCD of a a gives &! that is a a! The solutions are and of which is extraneous This procedure is summarized in: Theorem XV: Specifying solutions when there is one quadratic denominator Consider the rational equation - + /0 12 a + a To create specified solutions some of which may e extraneous let E F G e factorale with zeros < and < These will e the solutions of the rational equation For extraneous solutions set + or equal to < or < and choose the remaining coefficients to satisfy: -/E -E/ +/-01F One useful solution is 0 F1+/- +02G 2+0G While tedious for humans these rules are easily implemented y software p 23 Copyright!' y Alan Roeuck May not e reproduced without permission

24 Part IV) A rational equation with a quadratic numerator We consider the equation - / a+ a This equation is a modification of the form found in Theorems XIII through XV It includes one term with a quadratic numerator and a quadratic denominator Š The ninth coefficient is 4 rather than 3 so that software will not confuse it with È The reduced equation is a-/1 a+/-02+04! Therefore the reduced equation will e linear with a single rational solution iff -/1 that is the sum of the coefficients of in the numerators on the left side of the equation equals the coefficient of in the numerator of the right side Example & & (% reduces to with solution ( % & (!ß & Again it is easy for the instructor to create a formidale-looking rational equation which reduces to linear If -/1Á! the equation is quadratic with discriminant H a+/-02 % a-/1 a+04 The solutions will therefore e rational iff H is a rational numer squared Since the equation for H is linear in 1 and in 4 we can ensure rational solutions y choosing all non- 1 coefficients aritrarily choosing a tenth free parameter 5 to e a rational numer setting H 5 and then solving for 1: 1-/ a +/-02 5 %+04 a Alternately we can apply the aove procedure except that we choose all non- 4 coefficients and solve for 4: 4+0 a +/-02 5 % a-/1 Theorem XVI Method of free parameters for the quadratic-numerator equation - / The equation + a+ a has rational solutions if: I) 5 is any rational numer and 1 - / a +/-02 5 %+04 a other coefficients eing any rational numers or II) 5 is any rational numer and 4 +0 a +/-02 5 % a-/1 other coefficients eing any rational numers We could also apply the method of equal discriminants Start with a factorale quadratic E F G and set H a+/-02 % a-/1 a+04 F %EG therefore +/-02 F and a-/1 a+04 EG Choose rational values for all parameters except 2 and 4 and solve the first equation for 2 and the second for 4 p 24 Copyright!' y Alan Roeuck May not e reproduced without permission

25 Specifying solutions in advance If we want to specify solutions in advance we use the method of equal coefficients: Start with a factorale quadratic E F G The reduced equation is a-/1 a+/-02+04! So if this is to e the equation E F G! then the coefficients + through 4must satisfy: -/1E +/-02FßThe natural solution is: +04Gà Ú1-/E Û 2+/-0F Ü4+0G That is choose + through 0 aritrarily and then calculate a1ß 2ß 4 y the aove equations: Theorem XVII Specifying solutions for the quadratic-numerator equation Consider the quadratic equation E F G! with solution s < and < Choose + through 0 aritrarily and calculate 1-/E 2+/-0F 4+0G Then the equation - + / a+ a also has < and < as solutions These procedures are tedious for a human ut well-suited for implementation y software Example Let's create a rational equation of the aove form whose reduced equation is & a a& &! Solutions: ß Therefore aeß Fß G aß ß & Let's make + so that one denominator is This makes an extraneous solution Ú 1-/E-/ Û 2+/-0F /-0 Ü 4+0G 0& The remaining coefficients are aß -ß ß /ß 0 To avoid large values of 1 2 or 4 let's choose through 0 to each e! or : aß -ß ß /ß 0 aß ß!ßß Ú 1 Therefore Û 2 a! ' Ü 4!&' All coefficients: a+ß ß -ß ß /ß 0ß 1ß 2ß 4 a ß ß ß!ß ß ß ß &ß ( Equation: - / '' + a+ a that is & Reduced equation: &! that is a a&! Solutions are (extraneous) and p 25 Copyright!' y Alan Roeuck May not e reproduced without permission

26 Part V) Rational equations with three quadratic denominators Another type of rational equation common in elementary algera is (The author relented and used 3 for the ninth coefficient) / a+ a a a- a+ a- Note that the least common denominator is only cuic: a+ a a- As we did for the previous type of equation we will see how to create an equation of this form which reduces to any factorale quadratic equation you choose This makes it possile to produce extraneous solutions And if there is no need to specify the solutions ut only have them rational there are several other methods for creating such equations Note the naming scheme for the coefficients: +ß and - are the opposites of the zeros of the denominators in order from left to right and ß /ß 0ß 1ß 2ß and 3 are the coefficients of the numerators in order from left to right Reduced equation: a02 a+02-/13+13-/! Discriminant: H a+02-/13 % a02 a+13-/ Note that if 02! the reduced equation is linear and solutions are therefore rational 02! means that 20 that is the sum of the coefficients of in the numerators on the left side equals the coefficient of in the numerator on the right side & Example Equation: a a 1 a a a a Reduced: *%! What if the reduced equation is quadratic? In order to use the method of free parameters we would need the expression for H to contain at least one coefficient that occurs only linearly That's ecause we have to solve for that parameter without getting radical-containing solutions This does not occur here ut the method of equal discriminants does work: Therefore H a+02-/13 % a02 a+13-/ F %EG +02-/13 F and a02 a+13-/ EG Get Eß Fß G from a factorale trinomial choose all coefficients except two to e aritrary rational numers and then find the remaining two from these two equations ecause of the product +0ß the final two should not e + and 0 or else you will have a nonlinear equation ecause of 2 they should not e and 2 And so on So the final two could e + and or 1 and / and so on p 26 Copyright!' y Alan Roeuck May not e reproduced without permission

27 Specifying solutions in advance Choose any factorale quadratic equation and call it E F G! We will choose the coefficients + through 3 so that +02-/13F 02E / G The reduced equation a02 a+02-/13+13-/! will therefore e E F G!ß which will have rational solutions efore looking at the general method let's consider an example ( Start with a ( a & &! So solutions are & and ß and E ß F ß G & Therefore the coefficients must satisfy +02-/13 02 ß / & à Since there are only three equations with nine unknowns an infinite numer of rational solutions exist We can find them y assigning values to six of the unknowns and solving the equations for the remaining three Since for example the first equation contains the nonlinear factor +0 we don't want + and 0 to e two of the three unknowns we solve for ut note that all nonlinear terms contain +ß ß and - Therefore egin y aritrarily choosing +ß and - For example +ß & -% Now we have linear equations: 0&2%/13 02 ß 1 &3 %/ & à ecause of the second equation chose! and 0 : &2/13 2 ß Therefore 2 1 &3 %/ & à Now we have two equations: & a /13 /13' that is 1 &3 %/ & 1 &3 %/ & With three variales choose one aritrarily 1 &: /&3' /3 a that is & &3 %/ & %/ &3 & So /!Î and 3 Î Altogether + ß & - %! /!Î 0 1 & 2 3 Î that is that is / Equation: a+ a a a- a+ a-!î & Î a a& a& a% a a%! & a a& a& a% a a% p 27 Copyright!' y Alan Roeuck May not e reproduced without permission

28 The reduced equation is: ' *!& a & a ( a &! This is the quadratic ( equation with which we started so the solutions are and & We need to solve the following system for + through 3: General method +02-/13F 02Eß / G à where EFand Gare the given coefficients of the factorale quadratic E FG! Our method is to assign aritrary rational (in practice usually integral) values to six of the unknown coefficients + through 3 and then solve these equations for the remaining three ecause the equations are not linear in for example + and 0 we must choose wisely which three variales to solve for Since the second equation says 20E 2 ought to e one of the final three The other two could e / 1 or 3 The author chose to solve for a/ß 1ß 2 Transfer the terms not containing these to the right sides: 2/1F E0ß +1 -/ G 3 à Put the unknowns in the same order in each equation: /12F E0ß -/ +1 G 3 à Matrix notation: Ô Ô / Ô F-+03!! 1 E0 Õ- +! ØÕ2Ø Õ G 3 Ø Formal solution: Ô / Ô Ô F-+03 1!! E0 Õ2Ø Õ- +! Ø Õ G 3 Ø Calculate the inverse: It is proaly more convenient to leave the solution like this Note the following features: Ô / Ô + + Ô F Õ Ø - - E Õ! +-! Ø Õ G 3 Ø A unique solution only exists if +Á- And if +-identical so that the equation is not really of the form given the denominators on the left side of the rational equation are here / 1and 2are expressed as functions of EßFßGß+ through ß0ßand 3 If all coefficients other than / 1 and 2 are chosen to e integers then all the coefficients will e integers if + and - differ y so that + In the classroom we want to avoid coefficients that are too large Since the values of / 1 and 2 will involve the products +ß-ß-ß+0 and 3 the coefficients +ßß-ßß0and 3should e small numers p 28 Copyright!' y Alan Roeuck May not e reproduced without permission

29 Example Let's show that the aove algorithm reproduce our first example In that example +ß & - %! /!Î 0 1 & 2 3 Î Also E ß F ß G & We have a/ß 1ß 2 a!îß &ß Using the equations aove to calculate a/ß 1ß 2 we have: Ô + + Ô a& Ô! % & % % %! Õ! +-! Ø Õ! %! Ø Õ!! Ø Also Ô F Ô %! Î Ô!Î E0! Õ G 3 Ø Õ & a & a Î Ø Õ!Î Ø Therefore Ô / Ô! Ô!Î Ô!Î 1 Õ Ø %! & 2 Õ!! Ø Õ!Î Ø the correct values Õ Ø Theorem XVIII: Specifying solutions of the three-quadratic-denominators equation Consider the rational equation If 0 2 the reduced equation is linear and the solutions are rational / a+ a a a- a+ a- To make solutions rational when 0 Á2 use the following procedure: 1) Take any factorale quadratic equation E F G! The solutions of this equation will e the solutions of the rational equation Ñ Choose any rational numers +ß ß -ß ß 0ß and 3 with + Á - Ô / Ô + + Ô F-+03 Ñ Calculate / 1 and 2 as: E 0 +- Õ2 Ø Õ! +-! Ø Õ G 3 Ø %Ñ The reduced equation will then e the equation E F G! &Ñ Let the solutions of this equation e < and < If you choose + < then < will e an extraneous solution If you also choose < oth solutions will e extraneous 'Ñ To make the coefficients manageale oserve these guidelines: 3Ñ Choose integers and choose + and - to differ y 33Ñ Make sure that the products are not too ig Example Start with the polynomial a a& & So E &ß F ß G Next choose +%ß ß-&ßßwith 0and 3to e determined We need to evaluate the matrix Ô F-+03 Ô & %03 Ô %%03 E0 &0 %0 Õ G3 Ø Õ 3 Ø Õ 3 Ø Now choose 0 and 3 so that the aove three numers are not too ig: 0 ß 3 : Ô %%03 Ô %% Ô %0 % Õ 3 Ø Õ Ø Õ Ø p 29 Copyright!' y Alan Roeuck May not e reproduced without permission