sasogeek
  • 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 \(P = \frac{c}{a} \)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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hartnn
  • hartnn
u can use formula , right ?
hartnn
  • hartnn
\(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)
sasogeek
  • sasogeek
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....?

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sasogeek
  • sasogeek
and idk if i can use a formula... it wasn't stated. the question is just as i asked.
hartnn
  • hartnn
\(\(\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
sasogeek
  • sasogeek
ummm sorry, \(\huge x^2+\frac{b}{a}x+\frac{c}{a}\)
jiteshmeghwal9
  • jiteshmeghwal9
jiteshmeghwal9
  • jiteshmeghwal9
u acn find some tips by using this tutorial
sasogeek
  • sasogeek
@hartnn that's a lot of work lol
hartnn
  • hartnn
trust me, it isn't, simplification happens lot easily.....
sasogeek
  • sasogeek
ok i'll try :)
jiteshmeghwal9
  • jiteshmeghwal9
\[\huge{\frac{-b + \sqrt{b^2-4ac}}{2a}+\frac{-b - \sqrt{b^2-4ac}}{2a}}\]take the denominator common\[\LARGE{\frac{(-b+\sqrt{b^2-4ac})+(-b-\sqrt{b^2-4ac})}{2a}}\]
jiteshmeghwal9
  • jiteshmeghwal9
now by opening brackets u can solve :)
hartnn
  • hartnn
for product, u'll need this : \(\huge \color{red}{(a+b)(a-b)=a^2-b^2}\)
sasogeek
  • sasogeek
the radicals go to 0 and you get -2b/2a .... right?
hartnn
  • hartnn
yes.
hartnn
  • hartnn
wasn't that simple ?
hartnn
  • hartnn
'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
hartnn
  • hartnn
did u get the product ?

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