Help me find a good Algebra 1 book to supplement Saxon (not AoPS)

[stupidusername]

My 10-year old son just finished Saxon 8/7. (Hooray!) He is going to tackle Saxon Course 3 next. After that, Algebra.

 

We are very happy with Saxon, and will be using Saxon Algebra 1. However, I am considering supplementing the Saxon text with a second Algebra book, just to make sure he has a very strong foundation. He is young, so I see no need to rush things.

 

We tried Art of Problem Solving a few months ago. My son disliked it. I didn't like it much either because helping him required too much of my time.

 

I would like to try Jacobs or Dolciani, because so many people here have recommended them. Unfortunately, I cannot find an affordable copy of either book. Also, I need a solutions manual, and I cannot find an affordable version of those either. 

 

Advice?

[MEmama]

Have you checked eBay or Amazon for used copies? I think I paid around $35 for Jacobs and maybe another $15 for the TM. I was impatient or I would have waited for an even better price. :).

 

I feel like a broken record recommending Jacobs for younger algebra students, but we've been SO happy with it. It's worth the hunt and the money, IMO.

 

(that said, I've never used Saxon or supplemented math programs, so I don't know how that will work out for you. Jacobs is plenty on its own, for whatever that's worth).

[JeanM]

I am one of those who used Dolciani to supplement, but I never had a solutions manual. The answers (only answers, not worked solutions) to the odd numbered problems are in the back of the book, if that helps. If you want something to supplement that does have a solutions manual, you could consider Lial or Life of Fred.

[wapiti]

How about Alcumus? Or are you looking for supplemental instruction as well?

[stupidusername]

Have you checked eBay or Amazon for used copies? I think I paid around $35 for Jacobs and maybe another $15 for the TM. I was impatient or I would have waited for an even better price. :).

 

I feel like a broken record recommending Jacobs for younger algebra students, but we've been SO happy with it. It's worth the hunt and the money, IMO.

 

(that said, I've never used Saxon or supplemented math programs, so I don't know how that will work out for you. Jacobs is plenty on its own, for whatever that's worth).

 

Thanks for suggesting ebay. I hadn't thought of that. I just went there and was able to get a reasonably-priced copy of the Jacobs Algebra solutions manual. I got a used copy of the textbook from Amazon.com. Thanks again.

[EndOfOrdinary]

Half.com is eBay's version for books.  Same account long on, same feedback system, same whole deal.  They  just deal in media.  So if you do not find anything on eBay proper, try half.ebay.com and search there.  Ebay just bought out Half a few years ago, so they are only partially integrated.  There is no bidding with Half, just flat pricing and media mail.

[horsellian]

I haven't used the upper levels yet, but there's always MEP, which is free (a good price for a supplement!), and has the answers available.  It would probably take a bit more work to find the problems you need to match up with Saxon though.

[Momling]

I don't know about supplementing Saxon in particular, but we have used Zaccaro Problem solving genius, real world algebra, CWP 5&6, keys to algebra and word problems from hand on equations, plus Patty Paper geometry and Understanding geometry for pre-algebra/algebra supplementation. My guess though, is that Saxon provides plenty of opportunities to practice skills, so I'd focus on what you feel is missing from saxon.

[mathwonk]

Here is my favorite used books site, abebooks.com, and a link to several copies of Jacobs' Algebra for under $20:

 

http://www.abebooks.com/servlet/SearchResults?an=harold+jacobs&sts=t&tn=elementary+algebra

 

 

Correction:  see next few posts.

[wapiti]

Here is my favorite used books site, abebooks.com, and a link to several copies of Jacobs' Algebra for under $20:

 

http://www.abebooks.com/servlet/SearchResults?an=harold+jacobs&sts=t&tn=elementary+algebra

 

Unfortunately, unless I'm missing something, this may be for the test masters, not the text (see ISBN).

 

AFAIK, the cheapest on Abebooks is here http://www.abebooks.com/servlet/SearchResults?isbn=9780716710479&sts=t&x=66&y=12.  It's still cheaper than that on amazon right now http://www.amazon.com/Elementary-Algebra-Harold-R-Jacobs/dp/0716710471

[AimeeM]

Unfortunately, unless I'm missing something, this may be for the test masters, not the text (see ISBN).

 

AFAIK, the cheapest on Abebooks is here http://www.abebooks.com/servlet/SearchResults?isbn=9780716710479&sts=t&x=66&y=12.  It's still cheaper than that on amazon right now http://www.amazon.com/Elementary-Algebra-Harold-R-Jacobs/dp/0716710471

I think so. They are listed as paperback.

[mathwonk]

what about the next few listed books on abebooks, called the "Teacher's guides" for $25?  are they the actual book plus something?

 

no they are only about 250 pages against over 800 for the actual book.

[mathwonk]

wow, when i moved this spring i gave away literally thousands of dollars worth of books, at least if you have to buy them.  but you cannot sell them for anything.  i looked on powells books site once where they offered me about $5 for a great calculus book that they were themselves listing for sale at several hundred dollars.  i decided to just sell all my $50 and up books to local math grad students and profs for $5 each.  Dozens of them sold at that price in about two days.  The copies of Jacobs I just put out in the hall in front of my office for free.  Others i donated to the departmental math major library.  I of course miss many of them now, but you can't bring everything when you move.  I don't even have copies of many of my own research papers, and articles in books that i wrote myself.  i assumed they would be available online but i haven't found them all.  I am down from about 300 math books and hundreds of research papers, to about 100 math books and a handful of papers.  But I still have Euler's Elements of Algebra, Euclid's Elements (of geometry), and the works of Archimedes, and works of Gauss on number theory and differential geometry, Hilbert on foundations of geometry and a popular work "geometry and the imagination", and Riemann's collected works.  I recommend Euler as probably the best algebra book out there for a motivated person wanting to learn from an immortal genius who actually wrote his book for his butler.

 

Back when people learned algebra from Euler, they learned not just the quadratic formula, but also the cubic formula, which I myself only learned in graduate school.  Using Euler's clear explanation, I taught this material to 10 year olds at epsilon camp.  Almost no one learns this stuff in high school today.

[mathwonk]

heres a $49 copy of jacobs:

 

http://www.amazon.com/gp/offer-listing/0716710471/ref=sr_1_1_twi_1_olp?s=books&ie=UTF8&qid=1415753216&sr=1-1&keywords=harold+jacobs+algebra

[mathwonk]

for what it's worth here are my notes for epsilon camp introducing quadratic equations.  I immodestly suggest that if your student learns even what is presented below they are way ahead of a lot of algebra students. ( I am trying to present it Euler's way.)  Of course it may be worthless, and that is ok too.  Use whatever works, and keep searching until you find it.

 

 

This is the beginning of the notes for 2013:

Some of the problems for this summer involve solving quadratic, or second degree, equations.  If you have forgotten that skill, or are just learning it, I will present here an explanation I like, that makes it more clear, to me at least, than the usual one found in some books.  Please help me improve it by asking questions.  if you have a different approach that you like, please share it with us as well.

part I:

How the roots of a quadratic equation are related to the coefficients. 

Remark:  The "coefficients" of the quadratic expression X^2 -BX +C, are the letters (or numbers) -B and C.  We say -B is the coefficient of X and C is the constant coefficient.  Of course the coefficient of X^2 is 1 here.  I may be careless sometimes and call the coefficients B and C,  instead of -B and C.  So you need to look at the expression to see whether it is X^2+BX+C or X^2-BX+C.  I like using the form X^2-BX+C, because then B has a simpler meaning as we will see below, but technically, the word "coefficient of X" refers to whatever is in front of the letter X together with its sign.

To solve a quadratic equation means to find numbers X that satisfy an equation of form X^2 -BX +C = 0, where B and C are given numbers.  The numbers X which solve it are called "roots" of the equation, and B and C are called the coefficients (or technically -B and C).  E.g. we may want to solve for X such that X^2 - 20X +75 = 0.  Here the solutions are X =5 and X=15, as you can check by substituting X=5 and X=15 into the equation and simplifying.

The thing to notice is that the two coefficients in the equation, i.e. the numbers 20, and 75, are obtained by adding and multiplying the solutions together.  I.e. the coefficient 20 = 5+15, and the coefficient 75 = (5)(15).

This is a general principle.  E.g. to solve the equation X^2 - 19X + 34 = 0, we must find two numbers whose sum is 19 and whose product is 34.  Do you see what they should be?   [answer below]

The reason this principle holds is because every quadratic equation can be factored into a product of two factors using its roots.  In our first example, we have roots 5,15 for X^2-20X+75 =0, and the left side of the equation factors as X^2-20X+75 = (X-5)(X-15).  (Check it.)

In the second example the roots are X=2,17 (did you get these?), and the quadratic expression factors as X^2 -19X+34 = (X-2)(X-17).  (Check it.)

The "factor theorem" of algebra says that X=r and X=s are the roots of an equation like X^2 - BX+C =0, if and only if the quadratic expression factors as X^2 - BX+C  = (X-r)(X-s).

It is easy to see this in one direction, namely that whenever the factorization is true, i.e. whenever X^2 - BX+C = (X-r)(X-s), then setting X=r or X=s, does make the expression equal to zero.  (Setting X=r or X=s makes the right hand side equal to zero, so it must also make the left hand side equal to zero).

Let's assume the other direction as well, that if X=r and X=s are the roots of the equation X^2 - BX+C  = 0, then the factorization does hold: X^2 - BX+C = (X-r)(X-s).

Then what does this tell us about the relation between the roots, r,s and the coefficients B,C?  To see that, all we have to do is multiply out the right side of the equation
X^2 - BX+C = (X-r)(X-s).  This gives X^2 - BX+C = (X-r)(X-s) =X(X-s)-r(X-s) 
=X^2 -sX -rX +rs = X^2 -(r+s)X + rs. 

Now since X^2 - BX+C = X^2 -(r+s)X + rs, all the corresponding coefficients on the right and left must be equal, so B = (r+s) and C = rs.  Thus whatever the roots are for X^2 - BX+C = 0, they must add up to give B, and multiply out to give C. 

I.e. to solve X^2 - BX+C = 0 means finding two numbers r,s whose sum is B and whose product is C.  This gives a way to always find such numbers, as we explain next.

[mathwonk]

here are the first 4 algebra problems posed to the epsilon campers:  maybe one of your students would have fun with one or more of them.  they are harder than average though.

 

 

These three algebra problems are from the book Elements of Algebra by
Euler, the great 18th century mathematician.

1. Find a number such that if we multiply half of it by a third of it, and to the
product add half of it again, the result will be 30.

2. Find two numbers, the one being double the other, and such that, if we
add their sum to their product, we obtain 90.

3. A father leaves to his four sons $8,600 and , according to the will, the
share of the eldest is to be double that of the second, minus $100; the
second is to receive three times as much as the third, minus $200; and the
third is to receive four times as much as the fourth, minus $300. What are
the respective portions of these four sons?

here is a problem from an 1895 algebra book, Treatise on Algebra, by Charles Smith:

4. i) {a^2 (1/b - 1/c) + b^2 (1/c - 1/a) + c^2 (1/a - 1/b)} / {a(1/b - 1/c) + b(1/c - 1/a) + c(1/a - 1/b)}.
simplify.
Hint: the answer is a+b+c.

ii)  Related problem:  Factor this polynomial completely into linear factors:
{a^3(c-b ) + b^3(a-c) + c^3 (b-a)}

hint:  recall that if X = r is a root of a polynomial in X, then X-r is a factor.  What do you think happens if a=b is a "root"?

 

By the way, problem #4 from 1895, which is essentially undoable even by most teachers of today, gives you an idea of the decline in expectations in algebra over 100 years of textbook writing.  But this is the accelerated learners board.

 

By the way, if you are curious about the 1895 book Treatise on algebra by Charles Smith, do not buy the cheesy reprint by Forgotten Books, as it is an unreadably bad scanned copy.  There is no value in an algebra problem if the numbers are not legible.  This is a piece of junk as I learned to my chagrin.  I eventually bought an old library copy, but have since given it away.

[mathwonk]

here is what came "next" after post #15:

 

We have seen that the solutions p,q of the equation  X^2 -BX + C = 0, satisfy B = p+q, and C = pq, and we want to use this to find p and q, in terms of B and C.  

 

The trick discovered by the ancients was that this could be done if only we knew the value of p-q.  I.e. then we could add B to that value and get  B + (p-q) = (p+q)+(p-q) = 2p, and dividing by 2 solves for p.  Similarly subtracting p-q from B would give B - (p-q) = (p+q)-(p-q) = 2q, and we can divide by 2 and also get q.

 

Here is the trick:  notice that squaring (p-q) and squaring (p+q) gives almost the same thing, namely (p+q)^2 = p^2 + 2pq + q^2, and (p-q)^2 = p^2 - 2pq + q^2.  Thus these two squares differ by 4pq.  I.e.  (p+q)^2 - 4pq = (p-q)^2.  This says that the three quantities, (p+q)^2, (p-q^2), and pq, are related so that if you know any two of them you can find the third.  But we know two of them since we know B and C.

 

Thus,  (p-q)^2 = (p+q)^2 - 4pq = B^2 - 4C.  So taking a square root solves for (p-q) = sqrt(B^2-4C).  Thus 2p and 2q are obtained by adding and subtracting this from B, i.e. B ± sqrt(B^2-4C) = 2p, 2q, so p and q = (1/2)(B ± sqrt(B^2-4C)), the famous quadratic formula!

 

 

E.g. to solve  X^2 - 31X + 58 this way for p and q, we note that (p+q)^2 = (31)^2 = 96, and thus (p-q)^2 = 961 - 4(58) = 729.  So p-q = ± sqrt(729) = ± 27, and thus p,q = (1/2)(31 ± 27) = (1/2)(58) and (1/2)(4) = 29 and 2.

 

 

I love this because to me it explains the mysterious B^2-4C under the radical in the formula in a way I can relate to.