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Sagot :
We are given the following quadratic inequality
[tex]x^2+6x>16[/tex]Let us first convert it into quadratic form and then factorize the inequality
[tex]x^2_{}+6x-16>0[/tex]Now we need two numbers such that their sum is 6 and their product is -16
How about 8 and -2?
Sum = 8 - 2 = 6
Product = 8*-2 = 16
[tex]\begin{gathered} x^2_{}+6x-16>0 \\ (x+8)(x-2)>0_{} \end{gathered}[/tex]Now there are a few cases possible,
[tex]\begin{gathered} (x+8)>0\quad and\quad (x-2)>0_{} \\ x>-8\quad and\quad x>2_{} \end{gathered}[/tex]Since x > 2 meets the requirement of x > -8 so we take x > 2 from here.
[tex]\begin{gathered} (x+8)<0\quad and\quad (x-2)<0 \\ x<-8\quad and\quad x<2 \end{gathered}[/tex]Since x < -8 meets the requirement of x < 2 so we take x < -8 from here.
Therefore, the solution of the quadratic inequality is
[tex]x<-8\quad or\quad x>2[/tex]The endpoints are x = -8 and x = 2 (option A is correct)
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