Question

When factoring a quadratic equation in standard form ax²+bx+c = 0 and a is not greater than 1, the b and c share a special relationship.
To all; Share with the special relationship b and c share.

Answer 1

If it factors to (x+p)(x+q) then
b = p+q
c = p*q

Answer 2

whats the question?

Answer 3

nvm

Answer 4

(-2pb)^2/4p +k

Answer 5

What is the special relationship b & c share?

Answer 6

so a can be less than one but not greater?

Answer 7

to my understanding that is correct myininaya

Answer 8

hey i can attach something pretty cool about this if you give me a second ok?

Answer 9

like i found a new way to prove the quadratic formula and it might help you see the relation between a, b, and c

Answer 10

you just really need the first part ok?

Answer 11

but you can read the whole thing if you like

Answer 12

myininaya, this is sorta helpful, ia ma extremely math (retarded) and although I see lots that seems like it answers my question. can you put it "stupid" terms for me? all i need is the special relationship...

Answer 13

thats pretty good myininaya, i think i prefer the completing the square proof though :)

Answer 14

lol
okay lets see stupid terms:
what you want to do is find factors of a*c that add up to be b

Answer 15

find two factors of a*c that have product a*c and have sum b*

Answer 16

thanks cow :)

Answer 17

factoring a quadratic equation in standard form ax²+bx+c = 0 and a is not greater than 1, the b and c share a special relationship. so in order to find the "special relationship" you have to solve it first?

Answer 18

i think your question means if a=1
i see that it says a is not greater than 1
which means it could be less than 1 or equal to?
it doesn't make sense for it to exclude positive a and not negative a

Answer 19

\[x^2+bx+c\]
say the roots are
\[r_1 \& r_2\]
\[x^2+bx+c=(x-r_1)(x-r_2)=x^2-r_2x-r_1x+r_1r_2=x^2-(r_1+r_1)x+r_1r_2\]

Answer 20

i guess another question is: is there a standard "special relationship" between b & c in a quadratic equation OR does it vary by equation? perhaps i am overthinking this?

Answer 21

\[b=r_1+r_2 \& c=r_1r_2\]

Answer 22

and this is for a=1 okay?

Answer 23

i think what cow has is good

Answer 24

is there a way to put b in terms of c...is that what you are looking for

Answer 25

b equals the sum of factors of c...theres the special relationship (a=1)

Answer 26

\[ax^2+bx+c=a(x^2+\frac{b}{a}x+\frac{c}{a})\]
so we have roots \[r_1 \& r_2\]
=>
\[a(x^2+\frac{b}{a}x+\frac{c}{a})=a(x-r_1)(x-r_2)=a(x^2+-(r_1+r_2)+r_1r_2)\]
=>
\[\frac{b}{a}=-r_1+r_2 \& \frac{c}{a}=r_1r_2\]

Answer 27

oops there should be paranthesis around (r_1+r_2)

Answer 28

\[\frac{b}{a}=-(r_1+r_2) \& \frac{c}{a}=r_1r_2\]

Answer 29

there that looks better

Answer 30

=) i really appreciate your assistance!