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

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Answer 1 · 2018-04-04T11:46:41

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

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

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

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

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

Critical points:

$(x - 3) (2 x - 1) = 0$ $(x-3)(2x-1)=0$ . 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]

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