Here's the question you clicked on:
sasogeek
prove that the sum S and the product P of the quadratic equation \(ax^2+bx+c=0 \) are \(S= - \frac{b}{a} \) and \(P = \frac{c}{a} \)
u can use formula , right ?
\(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)
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....?
and idk if i can use a formula... it wasn't stated. the question is just as i asked.
\(\(\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
ummm sorry, \(\huge x^2+\frac{b}{a}x+\frac{c}{a}\)
http://openstudy.com/users/jiteshmeghwal9#/updates/500b81fce4b0549a892fa59c May this tutorial help you @sasogeek :)
u acn find some tips by using this tutorial
@hartnn that's a lot of work lol
trust me, it isn't, simplification happens lot easily.....
\[\huge{\frac{-b + \sqrt{b^2-4ac}}{2a}+\frac{-b - \sqrt{b^2-4ac}}{2a}}\]take the denominator common\[\LARGE{\frac{(-b+\sqrt{b^2-4ac})+(-b-\sqrt{b^2-4ac})}{2a}}\]
now by opening brackets u can solve :)
for product, u'll need this : \(\huge \color{red}{(a+b)(a-b)=a^2-b^2}\)
the radicals go to 0 and you get -2b/2a .... right?
'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