# Fixed Real Numbers...Part 1

• Dec 7th 2009, 06:18 AM
sologuitar
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.

• Dec 7th 2009, 07:37 AM
BabyMilo
Quote:

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.

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

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

Comparing coefficients gives

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

and

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

Hope this helps!
• Dec 7th 2009, 07:40 AM
Amer
Quote:

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.

$\displaystyle ax^2 + bx + c$

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

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 AM
HallsofIvy
Oh, this was a! I did the problems in the wrong order!
• Dec 8th 2009, 06:54 AM
sologuitar
ok...
Quote:

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

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

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

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

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

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

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

I would have not been able to work my way through this question.
• Dec 8th 2009, 06:57 AM
sologuitar
thanks...
Quote:

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

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

Comparing coefficients gives

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

and

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

Hope this helps!

This question is interestingly hard.