Question

prove that the sum S and the product P of the quadratic equation \(ax^2+bx+c=0 \) are \(S= - \frac{b}{a} \) and \(P = \frac{c}{a} \)

Answer 1

u can use formula , right ?

Answer 2

\(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 3

so i know that a quadratic equation is x^2+x(sum of roots)+product of roots
and you can have \(\large x^2+\frac{b}{a}+\frac{c}{a} \)
where from the negative with respect to the middle term....?

Answer 4

and idk if i can use a formula... it wasn't stated. the question is just as i asked.

Answer 5

\(\(\huge{x_1=\frac{-b + \sqrt{b^2-4ac}}{2a}}\\\(\huge{x_2=\frac{-b - \sqrt{b^2-4ac}}{2a}}\)\)\)
find x1+x2
and x1.x2

Answer 6

ummm sorry, \(\huge x^2+\frac{b}{a}x+\frac{c}{a}\)

Answer 7

http://openstudy.com/users/jiteshmeghwal9#/updates/500b81fce4b0549a892fa59c


May this tutorial help you @sasogeek :)

Answer 8

u acn find some tips by using this tutorial

Answer 9

@hartnn that's a lot of work lol

Answer 10

trust me, it isn't, simplification happens lot easily.....

Answer 11

ok i'll try :)

Answer 12

\[\huge{\frac{-b + \sqrt{b^2-4ac}}{2a}+\frac{-b - \sqrt{b^2-4ac}}{2a}}\]take the denominator common\[\LARGE{\frac{(-b+\sqrt{b^2-4ac})+(-b-\sqrt{b^2-4ac})}{2a}}\]

Answer 13

now by opening brackets u can solve :)

Answer 14

for product, u'll need this :
\(\huge \color{red}{(a+b)(a-b)=a^2-b^2}\)

Answer 15

the radicals go to 0 and you get -2b/2a .... right?

Answer 16

yes.

Answer 17

wasn't that simple ?

Answer 18

'so i know that a quadratic equation is x^2+x(sum of roots)+product of roots' <---NO
'so i know that a quadratic equation is x^2-x(sum of roots)+product of roots'<---YES

Answer 19

did u get the product ?