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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 \(P = \frac{c}{a} \)
 one year ago
 one year 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 \(P = \frac{c}{a} \)
 one year ago
 one year ago

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

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

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

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

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

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

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
ok i'll try :)
 one year 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\[\LARGE{\frac{(b+\sqrt{b^24ac})+(b\sqrt{b^24ac})}{2a}}\]
 one year ago

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

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

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

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

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