Question

50 pontos! Resolva a Inequação: −x2+3x+4>0 / / x2−3x−18≤0

Answer 1

Resposta:

-x^2+3x+4>0

DELTA=(3)^2-4(-1)(4)

DELTA=9+16

DELTA=25

(-3+_5)/-2

X'=-3+5/-2=2/-2=-1

X"=-3-5/-2=-8/-2=4

S={x€R/-1<x<4

x^2-3x-18<_0

Delta=(-3)^2-4(1)(-18)

Delta=9+72

Delta=81

(3+_9)/2

x'=6

x"=-3

S={x€R/-3<_x<_6}

${x pertencete \: aos \: reais \: tal \: que - 3 \leqslant x \leqslant 6}$

Answer 2

Explicação passo-a-passo:

a) $\sf -x^2+3x+4 > 0$

$\sf -x^2+3x+4=0$

$\sf \Delta=3^2-4\cdot(-1)\cdot4$

$\sf \Delta=9+16$

$\sf \Delta=25$

$\sf x=\dfrac{-3\pm\sqrt{25}}{2\cdot(-1)}=\dfrac{-3\pm5}{-2}$

$\sf x'=\dfrac{-3+5}{-2}~\Rightarrow~x'=\dfrac{2}{-2}~\Rightarrow~x'=-1$

$\sf x"=\dfrac{-3-5}{-2}~\Rightarrow~x"=\dfrac{-8}{-2}~\Rightarrow~x"=4$

Temos que:

• $\sf -x^2+3x+4 > 0$, para $\sf -1 < x < 4$

• $\sf -x^2+3x+4 < 0$, para $\sf x < -1~ou~x > 4$

• $\sf -x^2+3x+4=0$, para $\sf x=-1~ou~x=4$

Logo, $\sf S=\{x\in\mathbb{R}~|-1 < x < 4\}$

b) $\sf x^2-3x-18 \le 0$

$\sf x^2-3x-18=0$

$\sf \Delta=(-3)^2-4\cdot1\cdot(-18)$

$\sf \Delta=9+72$

$\sf \Delta=81$

$\sf x=\dfrac{-(-3)\pm\sqrt{81}}{2\cdot1}=\dfrac{3\pm9}{2}$

$\sf x'=\dfrac{3+9}{2}~\Rightarrow~x'=\dfrac{12}{2}~\Rightarrow~x'=6$

$\sf x"=\dfrac{3-9}{2}~\Rightarrow~x"=\dfrac{-6}{2}~\Rightarrow~x"=-3$

Temos que:

• $\sf x^2-3x-18 \ge 0$, para $\sf x \le -3~ou~x \ge 6$

• $\sf x^2-3x-18 \le 0$, para $\sf -3 \le x \le 6$

• $\sf x^2-3x-18=0$, para $\sf x=-3~ou~x=6$

Logo, $\sf S=\{x\in\mathbb{R}~|-3 \le x \le 6\}$