I was wondering......

Originally posted by joe shmo
I have yet another problem................. take a simple qaudratic......

ex: (x^2 -5x + 4)

It looks to me like the factoring can be solved algebraiclly

like: ab = 4

a + b = -5

except I usually get right back to square one ( another quadratic perhaps with different signs)

im trying to avoid that result............is there any way?😕

the best way to solve this type of system of equations algebraically (in my experience) is a method called "backsolving":

ex. (1) ab = 4
(2) a+b = -5

if you solve equation (1) for one variable in terms of another [i.e. a = b/4] and substitute that in to the other equation in PLACE of a it becomes:

(3) [b/4] + b = -5 this is easily solved algebraically.

5/4*b = -5 .... b = -4

then use either equation to figure out what a is (equation 2 is easiest: a + (-4) = -5 or a= -1)

Originally posted by Aetherael
the best way to solve this type of system of equations algebraically (in my experience) is a method called "backsolving":

ex. (1) ab = 4
(2) a+b = -5

if you solve equation (1) for one variable in terms of another [i.e. a = b/4] and substitute that in to the other equation in PLACE of a it becomes:

(3) [b/4] + b = -5 this is easil ...[text shortened]... en use either equation to figure out what a is (equation 2 is easiest: a + (-4) = -5 or a= -1)

that seemed good, but.......... ab=4 solved for b is a= 4/b

what happens when you plug 4/b in for (a)

4/b + b = -5

you get nowhere

Originally posted by joe shmo
that seemed good, but.......... ab=4 solved for b is a= 4/b

what happens when you plug 4/b in for (a)

4/b + b = -5

you get nowhere

you're entirely right... that was a strange but very lucky blunder on my part:

this type of system isn't as easily solved as i said - if you solve for one variable in terms of another like i said, you end up with a rewording of the original problem: (like your result: 4/b + b = -5 which when you multiply everyone by b, you get b^2 -5b +4 = 0 and we're back where we started)

the strategic use of noticing that c = product of the roots, and b = sum or roots in a quadratic equation is an alternate method to the sometimes messy or confusing quadratic formula.

so given ab = 4 and a+b=-5, you can generate a group of POSSIBLE integer values of a and b. in this example, test the factor pairs of 4: (1 and 4), (2 and 2), (-1 and -4), (-2 and -2) and look for a pair whose sum is -5.

this could prove useless if the coefficient of x^2 in the original problem is not 1, or if the solutions are irrational or imaginary numbers (this is often not the case in elementary problems, but is a very real possibility; such as in x^2 - x -1 = 0 where the roots end up as [1+sqrt(5)]/2 and [1-sqrt(5)]/2 aka the golden mean)

i hope this helps and isn't extra confusing... your best bet may be to use [-b +- sqrt(b^2-4ac)]/2a after all since it always spits out correct answers. but this is ugly to memorize, and often in easier problems is a major time sink.

Originally posted by Aetherael
you're entirely right... that was a strange but very lucky blunder on my part:

this type of system isn't as easily solved as i said - if you solve for one variable in terms of another like i said, you end up with a rewording of the original problem: (like your result: 4/b + b = -5 which when you multiply everyone by b, you get b^2 -5b +4 = 0 and we'r ...[text shortened]... ct answers. but this is ugly to memorize, and often in easier problems is a major time sink.

i don't have any problems solving quadratics........i was simply looking at the problem and assumed that ther was a way to determine algebraiclly the factors needed......this is just extracurricular math

I am interested to know how a problem like this would have been solved before they figured out to solve a quadratic

for instance....another guy suggested the answer lay in Diophantine equations......so i would like to know why he suggested that?

diophantine equations are a clever methodical procedure for finding a the solution space given more variables than equations (in this case using ab=4 leads to the quadratic we discussed, so we are looking at one equation in two variables)

the classic diophantine equation problem is an attempt to solve x^n + y^n = c, given integer valued x,y, and n (an example of this is when n=2 you are searching for "pythagorean triples"😉

in your problem's case, we are looking at a linear diophantine equation: ax + by = c

the elegance of the solution arrives when studying the greatest common divisor of a and b, comparing it to c, and discovering either infinitely many solutions or a distinct lack of solutions.

however, in your case, a and b are restricted to 1 (whereas they normally are allowed to vary infinitely). i.e. the number theoretic elegance is removed from the solving of the problem, and the limited "guess and check" work that is involved in a normal diophantine equation remains.

there is, however, a pretty beautiful construction of quadratic solutions though i don't know any good purely algebraic versions that avoid the quadratic formula.

some good reading on diophantine equations, if you're interested:
http://en.wikipedia.org/wiki/Diophantine_equation#Linear_Diophantine_equations (surprisingly informative)

http://mathforum.org/library/drmath/view/61325.html (a look at the process of solving a diophantine equation in a more apt situation)

Originally posted by Aetherael
diophantine equations are a clever methodical procedure for finding a the solution space given more variables than equations (in this case using ab=4 leads to the quadratic we discussed, so we are looking at one equation in two variables)

the classic diophantine equation problem is an attempt to solve x^n + y^n = c, given integer valued x,y, and n (an ...[text shortened]... 25.html (a look at the process of solving a diophantine equation in a more apt situation)

thanks a lot.....ill be sure to check the links out...... 🙂

any time... i love math problems, just discovered this forum on the site, been playing here forever lol also feel free to PM me or play some chess haha

Originally posted by joe shmo
i don't have any problems solving quadratics........i was simply looking at the problem and assumed that ther was a way to determine algebraiclly the factors needed......this is just extracurricular math

I am interested to know how a problem like this would have been solved before they figured out to solve a quadratic

for instance....another guy sugge ...[text shortened]... ted the answer lay in Diophantine equations......so i would like to know why he suggested that?

I only suggested Diophantine equations because you said you wanted integer solutions, which are the only permissible solutions for a Diophantine equation.

I doubt anyone did solve quadratics before they developed the quadratic formula, just because it's relatively simple to derive. The other methods are, to my eye, a little more in-depth.

The Newton-Raphson method is an iterative method that will provide you with a numerical answer. It works most of the time, but not all the time. The formula is:

x2 = x1 - y1 / f'(x1)

The algorithm involves choosing a point (x1, y1) on the curve, then taking the derivative, and computing the next point x2 using the above formula. For the next iteration, x2 becomes x1, and you calculate a new x2. A computer is handy for this solution. Here's a more detailed description:

http://en.wikipedia.org/wiki/Newton's_method

The rational coefficient method involves finding all the combinations of the factors of the first and last coefficients in the polynomial, then trial and error and some long division. For example:

2x^2 + x - 1 = 0

The factors of 2 are 1 and 2, and the only factor of 1 is 1. So you combine them by dividing the factors of 1 by the factors of 2, letting them vary between positive and negative, to get your list of possible roots:

+/- 1/2 = +/- 1/2
+/- 1/1 = +/- 1

When you find one that works, you do long division on the polynomial. Let's suppose I try -1 right off the bat 😉, here's what you get:

2(-1)^2 + (-1) - 1 = 2 - 1 - 1 = 0

OK, that works, so now we have to divide through by x + 1:

(2x^2 + x - 1) / (x + 1) = 2x - 1

Luckily, we end up with a linear equation after our first division, so we can solve it explicitly for x:

2x - 1 = 0

x = 1/2

Plugging this solution back into the equation, we see that it is indeed a solution:

2(1/2)^2 + (1/2) - 1 = 1/2 + 1/2 - 1 = 0

As you can see, this method is much more involved than using the quadratic formula, and it only works for rational coefficients. If the above example wasn't the clearest, here's a much better set of them:

http://en.wikipedia.org/wiki/Polynomial_long_division

ahhh synthetic division. good times. newton's method is a really nice look back after the advent of the calculus... very pretty. reminds me of euler's approximation: (y2-y1) = f'(x1)*(x2-x1) to predict a future height along a curve. note, that none of the quadratic methods described work out so well if the discriminant (b^2-4ac) is negative (except for the quadratic formula, OR if you learn to factor using complex numbers). cheers.