The Student Room Group

quadratic equation factorable

i'm wondering if there is a way to right away identify if a quadratic equation is not factorable without manually multiplying 'a' and 'c' and then checking if theres a factor combination of the product that adds up to give 'b'. [quad.]

thanks.
Sorry - I usually just work out possible combinations in my head, and if I can't find anything but I know that they are factorisable, then I use the ac method.
Reply 2
No real method I know of, I do what thegodofgod says usually...
If the discriminant is a square number (and all the coefficients are integers), then that means it's factorisable (well, that method's always worked for me, anyways).
Reply 4
The following equations you can use to find the solutions to quadratics (the + changes to a -):
(-b + (b^2 - 4ac)^1/2)/2a
(-b - (b^2 - 4ac)^1/2)/2a

If (b^2 - 4ac) is negative, then it doesn't factorise, as it is to the power of 1/2, which means a sqaure root. You cannot square root a negative number, meaning the quadratic doesn't factorise.
Original post by Revolution is my Name
If the discriminant is a square number (and all the coefficients are integers), then that means it's factorisable (well, that method's always worked for me, anyways).


This.

If the discriminants a square number then you're guaranteed rational solutions so the equation factorises.
Original post by Revolution is my Name
If the discriminant is a square number (and all the coefficients are integers), then that means it's factorisable (well, that method's always worked for me, anyways).


Hey - I never knew of that and just tried it out and it works :biggrin:

Cheers
Reply 7
"if discriminant is a square number, then equation factorises." brilliant. this would do. :biggrin: thanks guys.
i don't ever bother to try factorise things anymore unless i can see how it goes really easily. much quicker to whack it into the equation than to stare at it for a minute trying to work it out

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