Creating Good Equations for the Classroom: Rational, Logarithmic and Square Root aversion ""Î# 9Î#!"'

Transcription

1 Creating Good Equations for the Classroom: Rational, Logarithmic and Square Root aversion ""Î 9Î!"' Alan Roeuck Chaffey College We're math teachers. We don't just solve equations, we also create them. And the equations we create must e solvale y our students. If the equation is quadratic EB FB G!, it's not always easy to solve. And if the students have not een taught the quadratic formula or complex numers, it may e impossile for them to solve it. If the quadratic equation can e solved y factoring, the solutions will e rational numers, such as $ or. & What if it can't e solved y factoring? That is, what if the left side is unfactorale in the ordinary sense of the word? Then the quadratic formula is necessary, and solutions will e irrational or nonreal, & È"( & È& % ) 3 ) typically having the form or To e certain that the students can solve the equation we must sometimes ensure that the solutions are rational numers. That way the quadratic formula is unnecessary and the equation can e solved y factoring. Another thing: When the equation is rational, logarithmic or square root, extraneous solutions are possile. Therefore all solutions must e checked. And checking is generally a whole lot easier if the potential solutions are all rational. So we want rational solutions. Or at least, if we want rational solutions, then we can't just choose the equation parameters +ß, and - randomly. We must choose them wisely. It's easy to create a quadratic equation with rational solutions: Start with the two rational numer solutions you want, call them < and <, and another rational numer E, and multiply the expression " EB< a ab< " If the result is called EB FBG, then the solutions of EB FBG! are < and <. " But what if you must create a rational equation (one with B in a denominator)? Or a square root equation? Or a logarithmic equation? There do not seem to e any well-known methods for systematically creating equations like these that will have rational solutions. I have discovered several such methods: Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 1

2 Part I) Rational Equations My research on this topic egan when I had to write a test question of the following type: (Tussy and Gustafson, Elementary and Intermediate Algera, 5th Ed, p. )'!Ñ 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 complete the same jo when they work together. How long does it take Worker 1 to finish the jo, working alone? My prolem was to find values of + and, which make this word prolem relatively easy to solve. To answer the question you look at the rate of work for three situations: Worker 1 working alone, Worker 2 working alone, and the two workers working together. We have: " jo " " " > hours >+ together, V V V The equation is thus that is, V V V " Together " " " > >+, Multiply y the LCD: > a+, >+,! We could call this the reduced equation, defined to e the equation that results from clearing fractions. Apply the quadratic formula: +, È + %, +, È + %, > Þ That is, Solutions The equation is easiest to solve when the solutions are irrational numers such as $ È Þ rational numers, such as " & or ( rather than In this paper, we assume all equation parameters are rational numers. That's a given. But how can we choose the rational parameters + and, to make the solutions rational? There are two asic approaches to answering this question. Here's the first: Theoretical Method We're assuming that + and, are rational numers, preferaly whole numers. Therefore the solutions +, root. È + %, will e rational if, and only if, the square root part simplifies to something that is not a %* "" And that requires + %, a square, such as % or &, or even Þ But notice this: + %, + a,. We need + a, Square. Under what conditions is anumer aanother Numer Square? Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 2

3 If they're part of a Pythagorean Triple, such as a$ß %ß &, or a&ß "ß "$ If a+ß, a$ß % or a&ß ", the word prolem aove has rational solutions. Try it: +& and,", that is,,': Equation: " " " " " " > >+,, that is, > >& ' Clear the fractions: > "(> $!! The solutions are and "&. a>"& a>! If > then Worker 's time is >& $, which is invalid. Only >"& hours is a valid solution. But + and, don't have to e whole numers. We only need that + a, arational Numer Since & " "$, it's also true that & " "$, that is & & & " "$ a " Š & Š & " ' So we could have a+ß, Š "ß &. Since, &, the equation would e As you can check, the solutions are > ß $ &. " " & > >" ' These values of + and, do not give a valid solution to the word prolem. The time required to complete a jo cannot e negative. But they do give rational solutions of the equation, which is our general ojective. Here is the general result for the aove type of equation: We start with the classic definition of a Pythagorean Triple: Any triple of positive integers abß Cß Dfor which B C D Þ But for us, the numers don't have to e positive integers. They can e any rational numers. Therefore it makes sense to define: Def A generalized Pythagorean Triple is any triple of nonzero rational numers a! ß " ß for which! " And then we will have to refer to Def A standard Pythagorean Triple is a Pythagorean Triple of positive integers. Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 3

4 For our purpose we need: Def A Pythagorean Pair is the first two components of a generalized Pythagorean Triple. Therefore a!" ß is a Pythagorean Pair iff! " arational numer And it turns out that: Theorem a!" ß is a Pythagorean Pair iff a!" ß anonzero rational numer a Bß Cß where abß C are the first two components of a standard Pythagorean Triple. (In this paper we will omit most proofs, including this one.) That's what we did aove when we generalized from a+ß, a&ß" to a+ß, Š "ß : " & And it turns out that: " " & &ß " "ß & a Š Theorem The equation " " " > >+, has rational solutions for > iff a+ß, is a Pythagorean Pair. The theoretical prolem of finding + and, to make the solutions rational has een completely solved for this type of equation. But there's another approach to creating classroom prolems of the aove form: Empirical Method (Sustitution of parameters) " " " > >+, Rememer the equation:. Let's just pick a simple value of + : + ( Now the equation is " " " > >(, " " " * *(, Suppose we'd like the solution to e >* hours. So replace > with *: This equation is easy to solve for,. Especially if you have a TI-89., ") "" Therefore the equation is " " "" > >( ") "% "" According to my TI- )*, the solutions are and *. Since * was the solution we specified, we could define "" to e the conjugate of *, with respect to the equation " " " > >(,. We just forced one solution to e rational, ut the other solution also has to e rational. Since solutions +, È ß if one is rational the other must e. + %, Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 4 "%

5 And not only did we make the solutions rational, we specified one of them ahead of time, that is, efore creating the equation. " " " > >+, Naming our ojective is really a family of equations depending on two parameters + and,. It's a two-parameter family. We're trying to find values of + and, which make solutions rational. Def Given a family of equations, a rationalizer of solutions is a set of values of all the parameters which make the solutions of the equation rational numers. And what we're doing is: Def Rationalizing an equation afamily values which make the solutions rational. is finding one or more of its rationalizers, that is, parameter So the rationalizers of " " " > >+, are all a+ß, such that a+ß, is a Pythagorean Pair. The general goal we discuss here is rationalizing certain types of equation families. And a second ojective is to specify one or more of the solutions ahead of creating the equation. We remark in passing that Pythagorean Pairs are actually quite plentiful: Given any rational numer - there are an infinite numer of Pythagorean Pairs with first component -, and also an infinite numer of Pythagorean Pairs with second component -. 7 That's ecause if a!" ß is a Pythagorean Pair, meaning that! " arational, then so is 8 a!" ß for any nonzero natural numers 7and Therefore! a!" ß is a Pythagorean Pair with first component -, and " a!" ß is a Pythagorean Pair with second component -. Thus, for example, in the equation " " ", we can take + to e any rational numer. > >+, Another theoretical method Consider the equation family " " " B B+, This also comes from a word prolem. This is similar to the rate-of-work equation aove, ut it has a crucial difference: the variale terms are not on the same side of the equation. Multiply y LCD: + + %+, Solution: B B +B+,!Þ È For rational-numer solutions, we require H+ %+,arational We could use the method of Pythagorean Pairs: Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 5

6 + %+, + Š È +,! " where a!" ß is a Pythagorean pair. The details are a it more complicated here, ut it can e done. Or we could use sustitution of parameters: choose a rational value of + and a rational solution B, and solve for (rational,). But let's consider a new method: Method of equal discriminants $ Example Start with the equation B (B "& ab $ ab &!. Solutions are &ß Þ Definition Given a quadratic equation +B,B -!, the discriminant of the equation is H, %+- For the aove equation, the discriminant is H a ( % a "& %* % $! "'*. Also rememer: Theorem The equation +B,B -! has rational solutions iff H arational The aove equation has rational solutions, and H "'* "$. Next, recall that the equation " " " B B+, reduces to a quadratic equation whose discriminant is H+ %+,. We could refer to + %+, as the "discriminant of the rational equation:" Def Given a rational equation, the reduced equation is the equation that results from multiplying y the LCD. Def 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 " " " B B+, whose reduced equation also has H "'*. Therefore its solutions are rational. For the equation B (B "&!, we have H %* % $!. The "Method of Equal Discriminants" is to make 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, ( " " ( B B( $! If + (ß this gives the equation Þ Solutions are $ and "!, which are times the solutions of the equation with which we started ˆ $ & and. And that equation has a leading coefficient of. Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 6

7 " " ( If + (ßthis gives the equation B B( $! Þ Solutions are $ and "!Þ These are times the original solutions. Here is the general result, stated without proof: The Method of Equal Discriminants Let EB FBGe any factorale trinomial, with discriminant F % EGÞ If + and, are chosen so that EG F + %+, F %EG, that is, + F and, then " " " B B+, has rational solutions. " " -. B B-. -. In particular, if - and. are any nonzero rational numers, has - and. as its solutions. The method of equal discriminants, like the method of Pythagorean triples and the method of sustitution of parameters, can e applied to equation families other than the ones shown so far. Let's consider some other types of equations: Rational equation with one quadratic denominator That is, -B. /B0 1B2 B+ B, B+ B, a a Note that the denominator on the right is the product of the denominators on the left. Therefore PGH ab + ab,. For example, B" $B% B( B& B a B& a B $ È$ Solutions of the aove equation are 3. In Algera I (and maye in Algera II) we don't want solutions to e complex numers with irrational coefficients. So we had etter choose the parameters + through 2 wisely. Method of sustitution of parameters If all you need is rational solutions this is always the easiest method. Go ack to the equation family -B. /B0 1B2 B+ B, B+ B, a a. Assign aritrary values to all ut one of the parameters. So that the student will not have a difficult factorization to find the LCD, the unspecified parameter should e in the numerator. Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 7

8 For example, B$ %B" B2 B% B' a B% a B' If we want one solution to e, for example, B&, then the equation must e true when Bis replaced y &. Therefore 2 must solve the equation Solve it: &$ % &" &2. That is, ( " "!2 &% &' a&% a&' " "" "" 2 ''. Therefore the equation is Clear fractions: B$ %B" B'' B% B' ab% ab' B) a ab&!. Solutions: )ß& Þ We can also specify oth solutions efore we create the equation. Assign aritrary values to all ut two of the numerator parameters. For example, If we want & to e a solution, then B$ %B0 B2 B% B' B% B' a a &$ % &0 &2!0 "!2 &% &' a&% a&', that is, ( "" "" If also $ is to e a solution, then Ð$Ñ$ % Ð$Ñ0 Ð$Ñ2 * 0" $% $' 2' a$% a$', that is, ( $ " Ú ( So we have the linear system Û Ü!0 "!2 "" "" * 0" 2' ( $ " Since the system is linear, the solutions must e rational numers. These are not the solutions of the equation, they are the values of the unspecified parameters. But we do need these parameters to e rational. Using technology it's easy to solve: 0 $ 2 %%. B$ %B$ B%% B% B' ab% ab' Equation: As you can check, solutions are & and $Þ This method can e modified if the instructor needs to create an equation with one (or even two) extraneous solutions. These are numers which are not solutions of the equation ecause they make one or more of the denominators equal to zero, ut which are solutions to the equation that results when you clear fractions. Students need to e ale to handle this situation, so the instructor needs to assign such prolems. The calculations are more complex, ut here is the result: Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 8

9 + is an extraneous solution of -B. /B0 1B2 B+ B, B+ B, a a iff 2 a,+ a Choose values of all parameters except 2, and calculate 2 y the aove equation. Example Start with $B% &B( 'B2 B B"" a B a B"" Here, + ÞWe will choose 2so that + is an extraneous solution. This requires 2 a,+ a a""ðña%$ ÐÑ' ÐÑ "% Equation: $B% &B( 'B"% B B"" ab ab"" Clear fractions: % ab ab*! Solutions of the fraction-free equation are and. But is not a solution of the original equation ecause it makes two of the denominators to e zero. Here is another theoretical result aout this type of rational equation: Rational solutions for one quadratic denominator * Consider the equation -B. /B0 1B2 B+ B, B+ B, a a If -/! the reduced equation is linear and there is one rational solution. If -/Á! we must find eight parameters. To make solutions rational, choose the seven parameters +,,ß-ß.ß/ß0ßand 1to e any rational numers, and then set 2+0,. a+/,-.01 3, 3any rational numer. %-/ a Unlike the first type of equation we studied, this result does not give oth a necessary and sufficient condition for solutions to e rational. It only gives a sufficient condition. And the result is consideraly less elegant. To improve the elegance, we can simplify the aove theorem. If 3+/,-.01ß(and it can, ecause 3 can e any rational numer) then the fraction in the expression for 2 is zero, and we have: Simplified method for rational solutions with one quadratic denominator Consider -B. /B0 1B2 B+ B, B+ B, a a To make solutions rational, choose the parameters +,,ß-ß.ß/ß0ßand 1to e any rational numers (except that -/Á!Ñ, and set 2+0,.. +/,-.01 -/ The solutions are! and Þ Note: If all coefficients are integers, and -/ ", solutions are integers. Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 9

10 Although somewhat inelegant, these results can easily e implemented with software. Speaking of which: Creators of mathematics-educational software may need algorithms for generating good equations. The aove could e considered such algorithms; here is another: Specifying solutions for a quadratic-numerator equation Consider the quadratic equation EB FB G!, with so lutions < " and <. athat is, EB FBG EaB< ab< " Choose + through 0 aritrarily, and calculate 1-/E 2+/,-.0F 4+0,. G -B. /B0 1B 2B4 Then the equation also has < and < as solutions. B+ B, ab+ ab, " Here's what it means: Pick two rational numers and call them < " and <. Then choose another nonzero rational numer E, preferaly ", and calculate EaB< ab< ". Call the result EB FB G. So E, F and G are known. Next, choose any rational numers + through 0, and calculate 1ß 2 and 4y the aove equations. -B. /B0 1B 2B4 B+ B, ab+ ab, " Then the equation will have as solutions the numers < and < you started with. And note: If you choose + < " (and you can, ecause + can e any rational numer), then < " will e an extraneous solution: The equation will e -B. /B0 1B 2B4 B< B, ab< ab,, " " therefore two of the fractions will e undefined if B < " And if you choose + < " and, <, then oth solutions will e extraneous! The methods How did I discover these results? Without showing all the gruesome details, the key idea was this: Start with (for example) the equation: -B. /B0 1B2 B+ B, B+ B, a a Multiply y the LCD: a-/ B a+/,-.01b+0,.2! Now recall the discriminant of a quadratic equation: Equation: EB FBG!. Its discriminant: HF %EG Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 10

11 And then a quadratic equation has rational solutions iff Harational numer To make solutions rational, make the discriminant the square of a rational numer. Back to a-/ B a+/,-.01b+0,.2! Discriminant: H a+/,-.01 % a-/ a+0,.2 There are several ways of making H equal the square of a rational numer. We could, for example, use Pythagorean pairs. Or we could use the Method of free parameters To get the results aove called "rational solutions for one quadratic denominator," we do this: If Harational numer ßthen our equation has rational solutions. Call that rational numer 3. Then H3 that is, a+/,-.01 % a-/ a+0,.2 3 ( 3 is a "free" parameter ecause it does not occur in the equation.) Solve the aove equation for 2: 2 +0,. a +/,-.01 %-/ a And note that if the other parameters are all rational numers, so is 2. 3 We don't want an equation like È"( B& B' (B % B$ B% ab$ ab%, with an irrational coefficient. Method of equal coefficients To get the result aove called "specifying solutions for a quadratic-numerator equation," we do this: Equation: -B. /B0 1B 2B4 B+ B, ab+ ab, Multiply y LCD: a-/1b a+/,-.02b+0,.4! Now pick two rational numers < " and < which will e the solutions of the equation, and another rational numer E. Then calculate EaB< " ab< and call it EB FBG. So EßF and Gare known. We want our equation to e EB FB G!, for then the solutions will e < and <. " Since our equation is a-/1b a+/,-.02b+0,.4!, Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 11

12 -/1EÞ we make the two equations have equal coefficients: +/, F ß +0,. 4 G à Ú 1-/E The natural solution is: Û 2+/,-.0F This is the result given aove. Ü 4+0,.G The method of partial fraction expansion The TI- )* can write the partial fraction expansion of a rational expression. In the F menu, choose $ : expand. Or, at wolframalpha.com, write "partial fraction expansion of á " For example, partial fraction expansion of B& "(Î"$ *Î"$ ÐB'ÑÐB(Ñ B' B( We can use this technology to create rational equations with specified solutions, including one or more extraneous solutions: ab ab& ab ab'! Example Start with. When you clear fractions, the equation is abñðb&!. Solutions are and &, with extraneous. But we don't want to give it to the students in this form. It's too easy. "Expand" the numerator in this way: a B a B& B a B& a B& B a B& a B& ab ab' ab ab' ab ab' ab ab' BaB& ab& The equation is now ab ab' ab ab'!. Write the partial fraction expansion of the second fraction. This is possile ecause its numerator is of degree one less than its denominator: B&B ""Î (Î B B' B' B a a! B "!B ( "" B )B" B B' Times : This equation has solutions and &, with extraneous. Equations with two extraneous solutions, created with partial fractions ab ab& Example Begin with ab ab&!. When cleared of fractions, the equation is a B a B&!Þ Solutions are and &, oth extraneous. B$B"! ab ab& B Expand the numerator:. The numerator has the quadratic term. Therefore in the numerator, sutract and add a quadratic expression that egins with B : Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 12

13 ab ab& B %BB %B ab ab&! abe sure the quadratic expression is not a multiple of B or B& That is, a B a B& B %B B %B ab ab& ab ab&! B was chosen so that the first numerator will e of degree one: First numerator ab ab& B %B(B"! Since the numerator of the first fraction is of degree one less than its denominator, it can e expanded into partial fractions: First fraction (B"! ab ab& %&Î( B& %Î( B Equation: %&Î( B& %Î( B B %B B $B"!! %& % (B )B B& B B $B"! Times (: Times LCD: (B a ab&!. Solutions are ß &, oth extraneous! ab+ ab, A different way to make extraneous solutions Start with the identity ab+ ab, ". True for all B -values except + and,. Expand the numerator: B+,B+, a ab+ ab, " Add a+, B+, B a a a to each side: +, B+, B+ B, ab+ ab, "a B+ ab, This equation is also true for all B -values except + and, B a+, B+, ab+ ab, ab+ ab, Change the constant " to -: - ab+ ab, This is equivalent to ab+ ab, -. If -Á", there is no solution. If you multiply oth sides y the LCD and zero out one side, the result is So the solutions are ab+ ab, a"-! + and,, oth extraneous. We can generalize in one more way: Replace - with 0ÐBÑ. Then the reduced equation is ab+ ab, a"0ðbñ! Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 13

14 Since + and, are extraneous solutions, this equation is equivalent to 0ÐBÑ ". Therefore we have the following result: Theorem Consider the equation B a+, B+, ab+ ab, ab+ ab, 0ÐBÑ. If 0BÑ ( is of degree 8, the reduced equation is of degree 8Þ The extraneous solutions are + and,, and the genuine solutions are the solutions of 0ÐBÑ 1, if they are not + or,. Before giving the equation to your students, expand the fraction on the right side of the equation into partial fractions: Example Start with B ab ab$ $ B' a ab ab$ Expand into partial fractions: B B B$ *Î& %Î& B$ B a a Times &: &B * % B B$ "! B$ B a a Times LCD: &B &B $!!Þ Solutions: $ß, oth extraneous! Consider the form È +B, -B. Part II Square root equations?, + + Define a new variale:?+b,, so that B. The equation ecomes -,- È? +? +. Since the coefficients are aritrary, we might as well work with the form È? -?. Note that any valid solution must e non-negative, ecause È not a real numer. We want solutions to e rational numers. Therefore È? must e rational, for it equals -?.. Therefore solutions must e squares. Let's solve the equation: Square:?-? -.?. That is,!-? a-."?. Solutions:? "-. È "%-. - Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 14

15 Since we squared the equation, some of these solutions may e extraneous. Solutions will e rational iff the discriminant is the square of a rational numer. So let < e any rational numer, and set H<, that is, "%-.<. We can either choose - aritrarily and then solve for., or choose. aritrarily and solve for -. Since. is a term y itself in the original equation, the results are more elegant if we solve for.: " %-. < means. The square root equation is easier to solve if. is an integer. This is possile only if "< is a multiple of %, which only occurs if < is odd. This proves the following result: Let -Á! and < e aritrary rational numers. "< %- Rationalizing equations with one square root "< %- If. ß the equation È? -?. has rational solutions for?. If < is an odd integer,. "< %- can e an integer. The solutions are?š < " - ß possily extraneous. % Example If <& then "< %, and. %- To make. an integer, %- could e %ß ), " or 24. That is, - could e "ß ß $ or ' % 3) If - ', then. " % Equation: È? '?". Squared:? $'? "? ", that is, $'? "$? " a*? " a%? "! " " " " * % % * Solutions are ß Þ But is extraneous ecause it makes '?" negative. is the solution. % ) 33Ñ If - then. $ Equation: È??$. Squared:? %? "? *, that is, %? "$? * ab " a%b *!Þ * * % % Solutions are " and Þ But " is extraneous, so is the solution. Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 15

16 % % 333Ñ If - ", then. '. The equation is È?? ' Squared:?? "?$', that is,? "$?$' a?% a?*! Solutions are % and *, ut * is extraneous. The solution is %. Note that the specific equation È??$ gives us an unlimited numer of related equations, all with rational solutions: Just choose rational numers + and, and Example Let let? +B,Þ?B$. Equation: ÈB$ ab$ $ That is, È B $ %B * Squared: B $ "'B (B )" That is,! "'B (%B )%! )B $(B %!ab a)b" B! or )B"! Bß " ) Checking shows that only " ) is a solution of the original equation. The method of sustitution Let < e rational. If? < is a solution, then the equation is true when < is sustituted for?: Equation: È? -?. È Sustitute < : < -<., that is, l<l-<. But since the solution we chose is <, we may assume < is non-negative. Therefore < -<. If - is any rational numer, then. <-< is also rational Þ This proves (most of) the following Theorem If <! and -Á! are any rational numers, the equation È? -? <-< -<" - has the rational solutions?< and? Š Example Let < and - $. Then <-< $ %"% Equation: È? $? "%. Squared:? *? )%? "*' That is,! *? )&? "*' Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 16

17 Factor:!aB% a*b%* B%ß %* * -<" $ " ( %* - $ $ * Note that Š Š Š Check %: È %* % $ %"% Check : É %* %* $ "% * * * ( %* $ $ ( ( $ $ ""% "% Only % is a solution. What aout starting with the more general form È+B, -B.? negative. This way, solutions may e Assume < is a solution, and solve for,: È+<, -<. Square +<, a-<. Solve for,:, a-<. +< Theorem Consider the equation È +B, -B. Let <ß +ß - Á! and. e rational numers. If,a-<. +<, the equation has rational solutions: B<ß +- <-. - Example Choose <. We don't want a-<. +< to e large, so let -%ß.&ß+ Þ Then,a-<. +<a)& a a & Equation: Square: È B&%B& B & "'B %!B &! "'B %B!! ab a)b& B (extraneous), & ) Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 17

18 Two square roots The most general equation with two radicals containing linear B-terms in oth radicands would e, when simplified y setting?+b,, È?- È.?/0 An analysis similar to the one carried out aove shows necessary conditions for rational solutions: Equations with two square roots If the parameters -ß.ß 0 and < are rational, and -. Á ", then solutions of the equation -< 0 -." <.0 "-. È?- Ê.?Š 0 are?š rational, some possily extraneous. When the equation is squared twice, the result is -<0 -." a"-.? a- < 0?Š! Since the solutions are rational, this trinomial is factorale. To avoid unduly -<0 -." urdening students, make sure ¹ ¹ and k"-. k are small. Note this: Since - only occurs in the twice-squared equation as -, and 0 only occurs as 0, the eforeyou-check solutions will not e changed if you reverse the signs of - or 0. Example We must choose -ß.ß 0 and <. To make - < 0 small, let's make it a a $ (Þ So let - ß< and 0$ Then <.0 "-.. $ "a. %*. "%. We prefer. so that this will e an integer. The only possiilities are. and.!þ But if.!ß the equation is È <.0?- Ê0?Š 0, that is, È? -È constant 0 "-. This is too easy to solve to e worth analyzing here. Let's choose. ßso that %*. "%. <.0 Equation: È?- Ê.?Š 0ß that is, È? È "-.?$ Isolate larger radical: È? È?$ Square: % a??' È?* )?)?' È?* Isolate radical: ' È? "(? Square: $'? %*? "%? " Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 18

19 ! %*? &!? " B"ß " %* Both are extraneous. If we reverse the sign of -, the efore-checked solutions will e unchanged, and the equation is: This time, " %* is a solution. È? È?$ And, as efore, we can take the equation È? È?$ and generate an unlimited numer of different equations, all with rational solutions, y letting? +B, with +ß, rational numers. We can also consider the simpler case where -" so that the equation is È? È.?/ 0: Simplified two-square-root equation If the parameters.ß 0 and < are rational, and. Á ", then solutions of the equation < 0." <.0 ". È? Ê.?Š 0 are?š rational, some possily extraneous. When the equation is squared twice, the result is <0." a".? a< 0?Š! Since the solutions are rational, this trinomial is factorale. To avoid unduly <0." urdening students, make sure ¹ ¹ and k". k are small. Part III Logarithmic equations We consider the form log a+b - log a.b / 0,, As the ase of a logarithm,, must e positive ut not ". The other parameters can e any real numer, ut we will only consider rational parameters. And, of course, we want to choose the parameters so that the solutions will always e rational. Compress the logarithms into one logarithm: log a+b- a.b/ 0., Exponential version: a+b- a.b/, 0 We need to choose + through 0 so that the solutions for B are rational. The simplest method is: Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 19

20 The method of sustitution of parameters Example Choose values for all ut one of the equation parameters: ab $ ab / %. This corresponds to the logarithmic equation log ab$ log ab/ % % If B is a solution, then a a $ a/ % That is, a/ "' Solve for /: / "% So the exponential version of the logarithmic equation is Multiply, and get zero on the right: Factor: Solutions are B ß * ab $ ab "% "' B &B &)! ab ab*! The logarithmic equation is log ab$ log ab"%. The numer we chose,, is an % % extraneous solution ecause it gives us log a negative numer % But B * is a valid solution ecause it gives oth logarithmic expressions positive arguments. Note: The equation log ab-log ab/ 0, where the coefficients of B are oth ",,, always has one extraneous and one genuine solution. The more general equation can have two genuine solutions; an example is given near the end of this paper. General method Choose all equation parameters ut one. Also choose a rational value of B to e a solution of the original logarithmic equation. It might turn out to e an extraneous solutionþ Sustitute the chosen parameter values and the value of B into the equation a+b- a.b/, 0, and then solve for the unspecified parameter. Be sure the unspecified parameter is not, or 0, for then the solution for, or 0 will not necessarily e rational. Also, choose the solution and the specified parameters so that the unspecified parameter will not turn out to e too complicated a numer If, for example, / is the unspecified parameter and the solution is <, then we solve for / in the equation: Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 20

21 The result: a+<- a.</, 0 Method of Sustitution for rationalizing log a+b - log a.b / 0,, Exponential version: a+b- a.b/, 0 Choose rational values for all parameters except /, and choose a rational numer < such that +< - Á!. Solutions will e rational iff /.<, 0 +<- We prefer that / e an integer and not too large. Therefore choose +<- " and the other parameters so that l/l is not too large. +-.<+, -. Example Start y making 0 Note The solutions of the logarithmic equation will e < and +. a+<-. Usually one of these numers is extraneous., +<- +<- ": For example, %ß<$ß-"" Then /.< $.,. If, $ *, then /$.*. If. then / $ Equation: 0 a+b- a.b/,, that is, a%b"" ab$ * That is, )B "!B %! That is, ab$ a%b(! Solutions are B $ß ( % Since the original equation is log a%b "" log ab $ ß ( is extraneous. The solution is $ $ % B$Þ We conclude this section with Six easy formulas for creating logarithmic equations with rational solutions. I will derive two of them and simply state the other four. I) Consider the equation ab, ab"!. Solutions are, and ". If this is the exponential version of a logarithmic equation, Expand: B a," B,!, will e the extraneous solution. BB a a,", Take log ase,: log B log a B a," ",, Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 21

22 Theorem The equation log B log ab a," ",, has solution B", with, as an extraneous solution. Example If,"(, log B log a B"' " "( "( Exponential version: BaB "' "(Þ B "'B "(!Þ B "ß "( (extraneous.) - - " + II) Start with ab, a+b"!þ Solutions are, and. - - Expand: +B a"+, B,! - - B+B"+, a, Take log : log B log a+b"+, -,,, - Theorem The equation log B log a+b"+, - has solutions, and (extraneous.),, Example + &ß, $ß - : log B log a &B " & $ $ $ log B log a&b %% $ $ Exponential version: B&B%% a $ " & - - " + &B %%B *!: B *ß (extraneous) Theorem The equation log B log ab", " has solutions, and " (extraneous).,, Theorem The equation log B log ab a,- " has solutions -and, (extraneous.),-,- Theorem The equation log B log ab a, " - has solutions " and, (extraneous.),, - - Theorem The equation log B log ab", - has solutions, and " (extraneous.),, - - Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 22

23 The Fill-in-the-Blank Method Suppose you need a logarithmic or square root equation, you're in a hurry, and you don't have time to look up one of Professor Roeuck's theorems. No prolem: Example Suppose you want an equation of the form È+B, -B.. Let & e one solution. Then È+ &,- &. All you must do is fill in the following lanks, with any four numers which make the statement true: ÈÐ Ñ &Ð ÑÐ Ñ &Ð Ñ The radicand must e a square, so the left side could e: È( &" (There are, of course, infinitely many possiilities.) È( &" is ', so the equation ecomes 'Ð Ñ &Ð Ñ Fill in the lanks; there are infinitely many possiilities. Let's try this: ' &Ð%Ñ This gives us the equation È( &" &% Since & is B, the equation is È(B " B % Square: (B " %B "'B "'! %B $B "&!ab& a%b$ B&ß $ % aextraneous We can tweak the procedure. If È( &" were equal to 'Ðit doesn't) then 'Ð$Ñ &* That gives us the equation È( &"Ð$Ñ &* Since & is B, the equation is: È(B " $B * Square: (B " *B &%B )"! *B '"B )!!ab& a*b"' B&and "' * Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 23

24 But & is extraneous ecause it results in the equation È( &" ' By acting as if the root is negative, we make the solution we chose to e the extraneous solution. Fill-in-the-Blank Logarithmic Equations Let's create an equation of the form log a+b - log a.b / 0.,, Exponential version: a+b- a.b/, 0 Suppose, 0 % "'. Then a +B- a.b/ "' Suppose $ is one solution: a+ a$ - a. a$ / "' Therefore we must fill in these lanks: aò Ó a$ Ò Ó aö a$ Ö "' Fill in the lanks so that you get two factors of "'. For example, Ò Ó a$ Ò Ó ). Could e $ a$ a"( Ö a$ Ö. Could e " a$ a& So we have a$ a$ "( a" a$ & % Since $ is B, the equation is a$b"( ab& % Solutions are $ and $ $ Since the logarithmic equation is log a$b "( log ab & ß $ is extraneous. % % $ Fill-in-the-lank for square root equation with two radicals Let's create an equation of the form È+B,- È.B/0 Let $ e a solution. Then we have to fill in these lanks: Make the radicands to equal squares. For example ÈÐ Ñ $Ð ÑÐ ÑÈÐ Ñ $Ð Ñ Ð Ñ È$ $(Ð Ñ È" $"Ð Ñ That is, %Ð Ñ Ð Ñ The first lank is the value of. Keep it small or the equation will have large numers when squared - twice. So let's try % Ð Ñ Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 24

25 The lank on the right is ). Altogether we have È$ $( È" $") Since the second $ factor is B, the equation is È$B( ÈB") Isolate one radical: È$B() ÈB" Square: $B ( '% $ ÈB " % ab " $B(')%B$ ÈB" B'" $ ÈB" Next we would have to square each side, which will produce numers too large for most students. Let's adjust the parameters The prolem is the ) in the original equation. That's 0, the numer on the right side. So we have to decrease 0. We don't want 0! unless we want an equation that's very easy to solve. - and 0 are defined y %Ð Ñ Ð Ñ Do this %a" Therefore - " and 0. The revised equation is È$B(ÈB" Isolate one radical: È$B(ÈB" Square: $B(%% ÈB"B" B % ÈB " B" ÈB" Square again B B " % ab " Neither is extraneous. B B $! Therefore B $ß " Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 25

26 Picking a fractional solution It's easy if the fraction is small. For example, start with " " É% (- É $0 This means " will e a solution. This equation is $ - 0Þ So choose - "ß0". The finished equation is È%B ( ÈB $ " Isolate one radical: È È %B ( " B $ Square: %B(" ÈB$B$ Isolate the radical: B $ ÈB $ Square: %B "B * % ab $ That is, %B %B $! ab " ab $! Both are valid solutions. B " $ ß The B< method Oserve that if :! and ;!, the following equation is clearly true when B< : È: + ab< - È;. ab< :-; The parameters :ß ;ß +ß -ß. and < can e any rational numers, except that : and ; must e nonnegative. One solution is <. The other solution is B % $ + <+-.<%+- ; %+-:;-. <%-.:;%-.: +- a. This gruesome expression does tell us one important fact: The denominator shows that if +-.! there is no second solution. That means the equation ecomes linear when squared twice. Example Start with È $ ab" a" È% ab" a" %Þ " is a solution. That is, È È $B " B "% Isolate one root: È È $B " B "% Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 26

27 Square: $B " B "% % ÈB "% % Isolate radical: % ÈB "% "( B Square: "' ab "% B $%B )* That is,! B ''B '&!ab" ab'& B "ß '& If we sustitute '& for B, the equation ecomes È"*' È"%% "%" : False B" is indeed the only solution. The equation with one square root is also amenale to this method: We start with È: + ab< -ab< : If :! the equation is clearly true when B<. And the other solution is B - <+-: Rational equations - Let's see if we can use the B< method for equations of the form -B. /B0 1B2 B+ B, B+ B, a a. Add two fractions with denominators which are relatively prime. For example, To make an equation with < a solution, we write with the parameters $ $" & ( $& $+ ab< -ab< $"/ ab< &, ab< (. ab< c&,ðb<ñ dc(. ab< d + through / eing any rational numers. $ $" & ( $& When B<, the equation reduces to, which is true. Therefore < is a solution. And since the denominators are not zero when B<, < is not extraneous. Just pick convenient values of the parameters + through /. The most important guideline is to ensure that the denominator on the right, when expanded, can e factored without undue difficulty. Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 27

28 Example Let $ e the solution: $+ ab$ -ab$ $"/ ab$ &, ab$ (. ab$ c&,ðb$ñ dc(. ab$ d $+ ab$ -ab$ $"/ ab$ & ab$ (" ab$ c&ðb$ñ dc(" ab$ d Choose values of, and.: Now the LCD is ab"" ab% ab% ab"" ab $B%% Equation: $+ ab$ -ab$ $"/ ab$ B"" B% ab $B%% $ ab$ $ ab$ $" ab$ B"" B% ab $B%% Choose +, - and /: That is, B* $B"" B& B"" B% B $B%% a Simplified: B* $B"" B& B"" B% B $B%% Solutions: $ß "" If you have technology, the method of sustitution is easier: Choose all parameters ut one, choose a solution, and solve for the remaining parameter. If you don't have technology availale, the aove method is reasonale. Logarithmic equations Let's apply the B< method to the equation log a+b - log a.b / 0,, Exponential version a+b- a.b/, 0. Let, and 0 e rational numers (preferaly integers) with,! and, Á 1, and calculate, 0, which will e positive. 0 0 Let : and ; e positive factors of,. That is, :;,. This could also e the trivial factorization : ", ;, 0. Then for any rational numers + and., the equation log a:+ ab< log a;. ab< 0,, That is, c:+ ab< d e;. ab< f, 0 has B< as one solution: When < is sustituted for B, the equation ecomes :;, 0, which is true. Since and are positive, the arguments of the log functions are positive when, so is not an : ; B < < extraneous solution. Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 28

29 And since the equation is quadratic with rational coefficients, the other solution must also e rational. 0 Example Let, $ *Þ Then let : * and ; ". Let < % e a solution. Then c:+ ab< d e;. ab< f, 0 ecomes c*+ ab% d e". ab% f * Assign values to + and.: c*& ab% d e" ab% f * That is, a&b * ab * * This corresponds to the logarithmic equation log a&b * log ab * $ $ The exponential version is equivalent to "!B "!$B &! That's too difficult for most of our students. The iggest prolem is the numer * in the equation a&b* ab* *. Let's reduce the * y changing the value of + in the expression *+ ab%. We'll also change.: c*" ab% d e"$ ab% f * The logarithmic equation is log ab& a$b"$ * $B B&'! ab% a$b"%! B %ß "% $ a& B log a$b "$ $ $ In this case (as you can check) neither solution is extraneous! This is unusual; logarithmic equations in this form usually have one extraneous and one genuine solution. Note finally that if you don't mind a cuic equation, you could start with :::, " $ And then write the equation c: + ab< d c: + ab< d c: + ab< d, " " $ $ < will e one (non-extraneous) solution of the equation log a: + ab< log a: + ab< log a: + ab< 0 " " $ $,,, 0 0 Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 29

30 Astract You're writing a test. You need to create rational (or logarithmic or square root) equations which reduce to quadratic ut have rational solutions. No need to raid other textooks. Come learn aout theorems and rules for creating a wide range of equations with strictly rational solutions. Bio Alan Roeuck has degrees in Physics and Mathematics from UCLA and has een teaching community college mathematics for thirty years. He currently teaches at Chaffey College in the Inland Empire, specializing in algera, calculus and statistics. Copyright 2016 y Alan Roeuck. May not e reproduced without written permission. page 30