Question

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

Answer 1

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

Explanation:

`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]