How do I use a sign chart to solve $2x^2-7x> -3$ ?

Question

3589 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-04T11:46:41

Solution: $x <0.5 and x>3 or (-oo,0.5)uu(3,oo)$

$2x^2-7x> -3 or 2x^2-7x +3 > 0$ or

$2x^2-6x-x +3 > 0$ or

$2x(x-3)-1(x -3) > 0$ or

$(x-3)(2x-1) > 0$

Critical points:

$(x-3)(2x-1)=0$ . Critical points are $2x-1=0:. x= 1/2$

and $x-3=0:. x= 3:.$ critical points are $1/2,3$

Sign chart:

When $x< 1/2$ sign of $(x-3)(2x-1)$ is $(-) * (-) = (+) ; > 0$

When $1/2 < x < 3$ sign of $(x-3)(2x-1)$ is $(-) * (+) = (-) ; < 0$

When $x> 3$ sign of $(x-3)(2x-1)$ is $(+) * (+) = (+) ; > 0$

Solution: $x <0.5 and x>3 or (-oo,0.5)uu(3,oo)$ [Ans]

How do I use a sign chart to solve 2x^2-7x> -32x^2-7x> -3?