Fixed Real Numbers...Part 2

**sologuitar** · December 7th 2009, 06:20 AM

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).

**I-Think** · December 7th 2009, 04:36 PM

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 $s=\frac{-b-{\sqrt{b^2-4ac}}}{2a}<br />$
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

**sologuitar** · December 8th 2009, 07:00 AM

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 $s=\frac{-b-{\sqrt{b^2-4ac}}}{2a}<br />$
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.

Thread: Fixed Real Numbers...Part 2

LinkBack

Thread Tools

Display

Fixed Real Numbers...Part 2

well...

Similar Math Help Forum Discussions

real part of a function

Finding fixed numbers

Finding the Real/Imaginary Part of f(z), g(z)

Fixed Real Numbers...Part 3

Fixed Real Numbers...Part 1

Search Tags