## 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 $$P = \frac{c}{a}$$

1. hartnn

u can use formula , right ?

2. hartnn

$$\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$

3. 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....?

4. sasogeek

and idk if i can use a formula... it wasn't stated. the question is just as i asked.

5. 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

6. sasogeek

ummm sorry, $$\huge x^2+\frac{b}{a}x+\frac{c}{a}$$

7. jiteshmeghwal9

8. jiteshmeghwal9

u acn find some tips by using this tutorial

9. sasogeek

@hartnn that's a lot of work lol

10. hartnn

trust me, it isn't, simplification happens lot easily.....

11. sasogeek

ok i'll try :)

12. jiteshmeghwal9

$\huge{\frac{-b + \sqrt{b^2-4ac}}{2a}+\frac{-b - \sqrt{b^2-4ac}}{2a}}$take the denominator common$\LARGE{\frac{(-b+\sqrt{b^2-4ac})+(-b-\sqrt{b^2-4ac})}{2a}}$

13. jiteshmeghwal9

now by opening brackets u can solve :)

14. hartnn

for product, u'll need this : $$\huge \color{red}{(a+b)(a-b)=a^2-b^2}$$

15. sasogeek

the radicals go to 0 and you get -2b/2a .... right?

16. hartnn

yes.

17. hartnn

wasn't that simple ?

18. 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

19. hartnn

did u get the product ?