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

Dec 9, 2017

Solution: ${x}^{2} > 3 x - 3 + 0.75$ ,No critical point, $x | \phi$.

Explanation:

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

Discriminant $D = {b}^{2} - 4 a c = 9 - 12 = - 3$ . 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 $\left(x\right) = - \frac{b}{2 a} = \frac{3}{2} \cdot 1 = 1.5$

Veterx $\left(y\right) = {x}^{2} - 3 x + 3 = 0.75$

So Veterx is at $0.75 , 1.5$ So ${x}^{2} - 3 x + 3 > 0.75$ or

${x}^{2} > 3 x - 3 + 0.75$

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