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

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hartnn Group TitleBest ResponseYou've already chosen the best response.0
u can use formula , right ?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
\(\huge{x=\frac{b \pm \sqrt{b^24ac}}{2a}}\)
 2 years ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
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....?
 2 years ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
and idk if i can use a formula... it wasn't stated. the question is just as i asked.
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
\(\(\huge{x_1=\frac{b + \sqrt{b^24ac}}{2a}}\\\(\huge{x_2=\frac{b  \sqrt{b^24ac}}{2a}}\)\)\) find x1+x2 and x1.x2
 2 years ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
ummm sorry, \(\huge x^2+\frac{b}{a}x+\frac{c}{a}\)
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
http://openstudy.com/users/jiteshmeghwal9#/updates/500b81fce4b0549a892fa59c May this tutorial help you @sasogeek :)
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
u acn find some tips by using this tutorial
 2 years ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
@hartnn that's a lot of work lol
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
trust me, it isn't, simplification happens lot easily.....
 2 years ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
ok i'll try :)
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
\[\huge{\frac{b + \sqrt{b^24ac}}{2a}+\frac{b  \sqrt{b^24ac}}{2a}}\]take the denominator common\[\LARGE{\frac{(b+\sqrt{b^24ac})+(b\sqrt{b^24ac})}{2a}}\]
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
now by opening brackets u can solve :)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
for product, u'll need this : \(\huge \color{red}{(a+b)(ab)=a^2b^2}\)
 2 years ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
the radicals go to 0 and you get 2b/2a .... right?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
wasn't that simple ?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
'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^2x(sum of roots)+product of roots'<YES
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
did u get the product ?
 2 years ago
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