# Thread: Fixed Real Numbers...Part 2

1. ## Fixed Real Numbers...Part 2

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.

(b) Use part (a) to verify that ax^2 + bx + c =
a(x - r)(x - s).

2. ax^2 + bx + x = 0. Solutions: r,s
I think you mean
ax^2 + bx + c = 0.

$x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}$
Let $r=\frac{-b+{\sqrt{b^2-4ac}}}{2a}$
And $s=\frac{-b-{\sqrt{b^2-4ac}}}{2a}
$

Hence
$(x-r)(x-s)=x^2+\frac{b}{a}x+\frac{c}{a}=0$
$x^2-(r+s)x+rs=x^2+\frac{b}{a}x+\frac{c}{a}=0$

Hence $r+s=\frac{-b}{a}$ and $rs=\frac{c}{a}$

2nd part should fall into place now

3. ## well...

Originally Posted by I-Think
ax^2 + bx + x = 0. Solutions: r,s
I think you mean
ax^2 + bx + c = 0.

$x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}$
Let $r=\frac{-b+{\sqrt{b^2-4ac}}}{2a}$
And $s=\frac{-b-{\sqrt{b^2-4ac}}}{2a}
$

Hence
$(x-r)(x-s)=x^2+\frac{b}{a}x+\frac{c}{a}=0$
$x^2-(r+s)x+rs=x^2+\frac{b}{a}x+\frac{c}{a}=0$

Hence $r+s=\frac{-b}{a}$ and $rs=\frac{c}{a}$

2nd part should fall into place now
Part 2 does not fall into place but I will play with it later today.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}