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Algebra 15 Online
OpenStudy (anonymous):

When I'm solving Quadratic Functions what's another way to solve them without using x=-b+-SQRTb^2-4ac all over 2a?

OpenStudy (anonymous):

factoring them, then using the zero principle

OpenStudy (anonymous):

How do i do that?

OpenStudy (anonymous):

post up the problem and i'll shew you

OpenStudy (anonymous):

there is another methods but this method which you mentioned will be more easy... though if you are interested then send me the problem..

OpenStudy (anonymous):

I need part C, I used the other formula for part B.

OpenStudy (anonymous):

thanks, give me a sec..

OpenStudy (anonymous):

Okay.

OpenStudy (anonymous):

your not going to like my answer, because it'll require you to erase some of your work. use the quadratic equation on Part C, and use factoring on Part B so, you know how to do the quadratic equation, for Part C. So, i'll show you how to Factor for Part B factoring is FOIL in reverse we have 4x^2 - 12x + 5 set up my FOIL in reverse, 1 and 5 are the only factors of 5, ( - 1) ( -5) I need to get a 4x^2, so I will put in factors of 4, which are 2 and 2 (2x-1)(2x-5) now I FOIL what I made to check if it is correct (2x-1)(2x-5) = 4x^2 -10x-2x+5 I did it right, so it's factored now. next, we use the zero principle, its like this: if a*b = 0, then a=0 or b=0 or both =0 let a = (2x-1) let b = (2x-5) solve for x 2x-1 = 0 2x=1 x=1/2 solve for x 2x-5=0 2x=5 x=5/2 so x=1/2 and 5/2 any questions?

OpenStudy (anonymous):

x ^{2}+5x +4=0 x ^{2}+(4\times1)x +1=0 x ^{2}+4x +x +4=0 x (x+4)+1( x+4)=0 ( x+4)( x+1)=0 x=-4,-1

OpenStudy (anonymous):

Just one, I have to use the two different formulas so the FOIL one is for part B what do I use for part C... or is that what the zero principle one is?

OpenStudy (anonymous):

@DemolisionWolf

OpenStudy (anonymous):

oh, so the FOIL and the Zero Principle are methods that have to be used together to solve for X. and for the other problme, use the quadratic equation

OpenStudy (anonymous):

Ooh, okay. Thank you! :D

OpenStudy (anonymous):

^_^

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