How do you solve $z^2-5z-7<0$ using a sign chart?

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Answer 1 · 2017-12-27T12:56:26

Solution: $-1.14 < z < 6.14$

$z^2-5z-7=0 or z^2-5z=7$ or

$(z^2-5z+2.5^2)=7+2.5^2 or (z-2.5)^2=13.25$ or

$(z-2.5)=+-sqrt 13.25 or z= 2.5+-sqrt13.25$

$:. z~~ 6.14 , z ~~ -1.14 :.$

$z^2-5z-7 <0 or (z-6.14)(z+1.14)<0$

Critical numbers are $z~~ -1.14 , z ~~ 6.14$

Sign chart:

When $z< -1.14$ sign of $(z-6.14)(z+1.14)$ is $(-) * (-) = (+) ; >0$

When $-1.14 < z < 6.14$ sign of $(z-6.14)(z+1.14)$ is $(-) * (+) = (-) ; < 0$

When $z > 6.14$ sign of $(z-6.14)(z+1.14)$ is $(+) * (+) = (+) ; > 0$

Solution: $-1.14 < z < 6.14$

graph{x^2-5x-7 [-40, 40, -20, 20]} [Ans]

How do you solve z^2-5z-7<0z^2-5z-7<0 using a sign chart?