Suppose a, b, and c are fixed real numbers such that

b^2 - 4ac ≥ 0. Let r and s be the solutions of

ax^2 + bx + x = 0.

(a) Use the quadratic formula to show that r + s = -b/a and

rs = c/a.

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- Dec 7th 2009, 06:18 AMsologuitarFixed Real Numbers...Part 1
Suppose a, b, and c are fixed real numbers such that

b^2 - 4ac ≥ 0. Let r and s be the solutions of

ax^2 + bx + x = 0.

(a) Use the quadratic formula to show that r + s = -b/a and

rs = c/a.

- Dec 7th 2009, 07:37 AMBabyMilo
- Dec 7th 2009, 07:40 AMAmer

$\displaystyle ax^2 + bx + c $

$\displaystyle a\left( x^2 + \frac{b}{a} \cdot x + \frac{c}{a} \right) $

r,s are roots of the equation so we can write the equation like this

$\displaystyle (x-r)(x-s)= x^2 + \frac{b}{a} \cdot x + \frac{c}{a} $

$\displaystyle x^2 +x(-r-s) + rs = x^2 + \frac{b}{a} \cdot x + \frac{c}{a}$

$\displaystyle -r-s = \frac{b}{a} \Rightarrow r+s = \frac{-b}{a} $

$\displaystyle rs = \frac{c}{a} $ - Dec 8th 2009, 01:57 AMHallsofIvy
Oh, this was a! I did the problems in the wrong order!

- Dec 8th 2009, 06:54 AMsologuitarok...
- Dec 8th 2009, 06:57 AMsologuitarthanks...