# [SOLVED] Sign of a Quadratic

• Apr 7th 2005, 04:01 PM
Brian D
[SOLVED] Sign of a Quadratic
Solve : x^2 - 4x + 3 < 0

I need to graph the solution also
• Apr 8th 2005, 02:34 AM
paultwang
Quote:

Originally Posted by Brian D
Solve : x^2 - 4x + 3 < 0

I need to graph the solution also

(1) Factorize
x^2 - 4x + 3 = (x-3)*(x-1)

(2) Think, when is (x-3)*(x-1) negative (less than zero)? When each parenthetical group has a different sign.

Possibilities:
(a) x<3 making (x-3) negative AND x>1 making (x-1) positive
(b) x>3 making (x-3) positive AND x<1 making (x-1) negative

Therefore, 1<x<3

On a number line, open circles at 1 and 3 and connect them.
• Apr 21st 2005, 06:47 AM
HallsofIvy
The simplest way to solve "complicated" inequalities is to solve the EQUATION. Where something is EQUAL to 0 separates where it is positive from where it is negative! (If your inequality involves fractions, you want to check points where the denominator is 0 also.)

x^2- 4x+ 3= (x- 3)(x- 1)= 0 when x= 3 or x= 1.

Now check a value in each of the three intervals x< 1, 1< x< 3, 3< x.

0 is less than 1 and 0^2- 4(0)+ 3= 3 which is positive.

2 is between 1 and 3 and 2^2- 4(2)+ 3= 4- 8+ 3= -1 which is negative

4 is larger than 3 and 4^2- 4(4)+ 3= 16- 16+ 3= 3 which is positive.

We now know that x^2- 4x+ 3 is negative for ALL numbers between 1 and 3 and negative for ALL numbers less than 1 or larger than 3.
• Apr 21st 2005, 10:45 AM
ticbol
To HallsofIvy
You are wrong!

"...and negative for ALL numbers less than 1 or larger than 3."
Yeah?
• Apr 21st 2005, 10:51 AM
MathMan
Quote:

Originally Posted by ticbol
You are wrong!

"...and negative for ALL numbers less than 1 or larger than 3."
Yeah?

Lets try and keep this place civil... :)
We all make mistakes
• Apr 21st 2005, 12:17 PM
ticbol
To MathMan
I did not start it.

HallsofIvy blew in here like he could tell me something.