How do you solve #x^2<3x-3# using a sign chart?

Question

1424 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-09T12:51:27

Solution: #x^2>3x-3+0.75# ,No critical point, #x| phi#.

#x^2 < 3x-3 or x^2-3x+3 < 0 ; a=1 , b= -3 , c=3#

Discriminant #D= b^2-4ac=9-12 = -3 # . Since #D#

is negative the roots are imaginary ,there is no critical

point. Since #a# is positive the parabola opens upward.

and vertex is minimumm point .

Veterx #(x) = -b/(2a)= 3/2*1=1.5#

Veterx #(y) = x^2-3x+3= 0.75 #

So Veterx is at # 0.75, 1.5# So #x^2-3x+3 > 0.75# or

#x^2>3x-3+0.75#

Solution: No critical point, #x| phi#.
graph{x^2-3x+3 [-10, 10, -5, 5]} [Ans]