- #1
bairdos
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Hi! I came across the below thread where a user ('krackers') asked for a proof of the PSF factorising method for quadratic equations.
The thread is now closed so I'd like to post my proof here.
(The proof considers the simplest case where there are no common factors for a,b,c in the quadratic. A proof for cases with a common factor can easily be created with the same structure as below with a common factor 'd' added.)
https://www.physicsforums.com/showthread.php?t=621835
Given a polynomial ax^2 + bx + c = 0 there are rational roots if (b^2 - 4ac) is 0 or a positive square number
The rational factors can be written as p/q and m/n (fractions in simplest form)
therefore the polynomial can be written as (x-p/q)(x-m/n) = 0
multiplying by qn: (qx-p)(nx-m) = 0
which becomes nqx^2 - mqx -npx + mp = 0
so a = nq, c =mp and b is the sum of the middle terms (-mq and -np)
note that a*c= mnpq and that the product of the two numbers (-mq) and (-np) is also mnpq
so therefore if the polynomial has rational roots then there must be two numbers (-mq and -np) that add to give 'b' and multiply to give the product of 'a' and 'c'.
These numbers can be found by trial and error (because m,n,p,q are all integers)
To prove that there is only and only one pair of numbers S,T you can solve the simultaneous equations: S + T = b, ST = ac
(using the quadratic formula and the condition that (b^2-4ac) is 0 or a positive square number)
(of course there can only be one pair of numbers because otherwise there would be more than one factorisation of the polynomial resulting in there being multiple possible sets of roots).
I developed this proof because I felt it was my obligation as a high school maths tutor to understand why everything that I teach works :)
The thread is now closed so I'd like to post my proof here.
(The proof considers the simplest case where there are no common factors for a,b,c in the quadratic. A proof for cases with a common factor can easily be created with the same structure as below with a common factor 'd' added.)
https://www.physicsforums.com/showthread.php?t=621835
Given a polynomial ax^2 + bx + c = 0 there are rational roots if (b^2 - 4ac) is 0 or a positive square number
The rational factors can be written as p/q and m/n (fractions in simplest form)
therefore the polynomial can be written as (x-p/q)(x-m/n) = 0
multiplying by qn: (qx-p)(nx-m) = 0
which becomes nqx^2 - mqx -npx + mp = 0
so a = nq, c =mp and b is the sum of the middle terms (-mq and -np)
note that a*c= mnpq and that the product of the two numbers (-mq) and (-np) is also mnpq
so therefore if the polynomial has rational roots then there must be two numbers (-mq and -np) that add to give 'b' and multiply to give the product of 'a' and 'c'.
These numbers can be found by trial and error (because m,n,p,q are all integers)
To prove that there is only and only one pair of numbers S,T you can solve the simultaneous equations: S + T = b, ST = ac
(using the quadratic formula and the condition that (b^2-4ac) is 0 or a positive square number)
(of course there can only be one pair of numbers because otherwise there would be more than one factorisation of the polynomial resulting in there being multiple possible sets of roots).
I developed this proof because I felt it was my obligation as a high school maths tutor to understand why everything that I teach works :)