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

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

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

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0so 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....?

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

hartnn
 3 years ago
Best 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

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

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

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

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

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

jiteshmeghwal9
 3 years ago
Best 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\[\LARGE{\frac{(b+\sqrt{b^24ac})+(b\sqrt{b^24ac})}{2a}}\]

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

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

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

hartnn
 3 years ago
Best 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
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