# Thread: Fixed Real Numbers...Part 1

1. ## Fixed 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.

2. Originally Posted by sologuitar
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.
$ax^2+bx+c = a(x-R)(x-S)$

$
\Rightarrow ax^2+bx+c = ax^2-a(R+S)x+aRS$

Comparing coefficients gives

$b = -a(R+S) \rightarrow R+S = \frac{-b}{a}$

and

$c = aRS \rightarrow RS = \frac{c}{a}$

Hope this helps!

3. Originally Posted by sologuitar
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.

$ax^2 + bx + c$

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

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

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

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

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

$rs = \frac{c}{a}$

4. Oh, this was a! I did the problems in the wrong order!

5. ## ok...

Originally Posted by Amer
$ax^2 + bx + c$

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

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

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

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

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

$rs = \frac{c}{a}$
I would have not been able to work my way through this question.

6. ## thanks...

Originally Posted by BabyMilo
$ax^2+bx+c = a(x-R)(x-S)$

$
\Rightarrow ax^2+bx+c = ax^2-a(R+S)x+aRS$

Comparing coefficients gives

$b = -a(R+S) \rightarrow R+S = \frac{-b}{a}$

and

$c = aRS \rightarrow RS = \frac{c}{a}$

Hope this helps!
This question is interestingly hard.