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Given the equation
[tex]y = x(x + 2) - 15[/tex]
which of the following inequalities gives all possible true values of x, given that
[tex]y < 0[/tex]

Answer choices:

a)
[tex]x > 3[/tex]
b)
[tex]0 < x < 3[/tex]
c)
[tex] - 5 \leqslant x \leqslant 3[/tex]
d)
[tex] - 5 < x < 3[/tex]
​

Respuesta :

kamilmatematik100504

Answer:   [tex]\Large \boldsymbol{d) -5<x<3}[/tex]

Explanation:

[tex]\displaystyle \Large \boldsymbol {} y=x(x+2)-15 =0 \\\\x^2+2x-15<0 \\\\\left \{ {{x_1+x_2=-2} \atop {\!\!\!\!\!\!x_1x_2=-15}} \right.=> x_1=-5 \ ; \ x_2=3 \\\\\\(x+5)(x-3) <0\\\\\\signs : +++ (-5) --- (3) +++[/tex]

                                  //////////  

[tex]\Large \boldsymbol{} x \in (-5 \ ; \ 3) \ \ or \ \ -5<x<3[/tex]

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