Question

if x^2+6x-27<0; x^2+3x+4>0 then x lies in the interval​

Answer 1

Answer:

$(a) \: x \in ( 3, 4)$

Step by step explanation :

${x}^{2} + 6x - 27 > 0$

$\implies (x + 9)(x - 3) > 0$

$\implies x \in ( - \infty, - 9) \cup( 3,\infty )$

Further,

$- {x}^{2} + 3x + 4 > 0$

$\implies {x}^{2} - 3x - 4 < 0$

$\implies (x - 4)(x + 1) < 0$

$\implies x \in ( - 1, 4)$

Taking intersection,

$x \: lies \: in \: ( 3,4)$