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Sagot :
Solution:
B. [tex](-6-\sqrt{29},-6+\sqrt{29})[/tex]
Hope this helps! If so, lmk! Thanks and good luck!
Answer: (-6-[tex]\sqrt{29}[/tex], -6+[tex]\sqrt{29}[/tex]) ==> D
Step-by-step explanation:
x^2 + 12x +7 < 0
x^2+12x+36-29<0
(x+6)^2-29<0
(x+6)^2<29
x+6<[tex]\sqrt{29}[/tex]
x<-6+[tex]\sqrt{29}[/tex]
x+6>-[tex]\sqrt{29}[/tex]
x>-6-[tex]\sqrt{29}[/tex]
(-6-[tex]\sqrt{29}[/tex], -6+[tex]\sqrt{29}[/tex]) ==> D
-6-[tex]\sqrt{29}[/tex] and -6+[tex]\sqrt{29}[/tex] aren't included in the solution since if these values are plugged in to x^2 + 12x +7, the expression will equal 0. That's not supposed to happen. The expression is supposed to be LESS than 0.
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