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

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Answer 1 · 2017-12-09T12:51:27

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

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

Discriminant $D = b^{2} - 4 a c = 9 - 12 = - 3$ $D= b^2-4ac=9-12 = -3$ . Since $D$ $D$

is negative the roots are imaginary ,there is no critical

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

and vertex is minimumm point .

Veterx $(x) = - \frac{b}{2 a} = \frac{3}{2} \cdot 1 = 1.5$ $(x) = -b/(2a)= 3/2*1=1.5$

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

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

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

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

How do you solve x2<3x−3x^2<3x-3 using a sign chart?