Question

Quanto é: |x² + x -3| > 2

Alguém me ajuda?

Answer 1

Explicação passo-a-passo:

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

$\sf \Delta=1^2-4\cdot1\cdot(-3)$

$\sf \Delta=1+12$

$\sf \Delta=13$

$\sf x=\dfrac{-1\pm\sqrt{13}}{2\cdot1}=\dfrac{-1\pm\sqrt{13}}{2}$

$\sf x'=\dfrac{-1+\sqrt{13}}{2}~\approx1,30$

$\sf x"=\dfrac{-1-\sqrt{13}}{2}~\approx-2,30$

$\sf |~x^2+x-3~| > 2$

Há duas possibilidades:

1)

$\sf x^2+x-3 > 2,~para~x < -2,30~ou~x > 1,30$

$\sf x^2+x-3-2 > 0$

$\sf x^2+x-5 > 0$

$\sf \Delta=1^2-4\cdot1\cdot(-5)$

$\sf \Delta=1+20$

$\sf \Delta=21$

$\sf x=\dfrac{-1\pm\sqrt{21}}{2\cdot1}=\dfrac{-1\pm\sqrt{21}}{2}$

$\sf x'=\dfrac{-1+\sqrt{21}}{2}~\approx~1,79$

$\sf x'=\dfrac{-1-\sqrt{21}}{2}~\approx~-2,79$

Temos $\sf x < \dfrac{-1-\sqrt{21}}{2}~ou~x > \dfrac{-1+\sqrt{21}}{2}$

2)

$\sf x^2+x-3 < -2,~para~-2,30 < x < 1,30$

$\sf x^2+x-3+2 < 0$

$\sf x^2+x-1 < 0$

$\sf \Delta=1^2-4\cdot1\cdot(-1)$

$\sf \Delta=1+4$

$\sf \Delta=5$

$\sf x=\dfrac{-1\pm\sqrt{5}}{2\cdot1}=\dfrac{-1\pm\sqrt{5}}{2}$

$\sf x'=\dfrac{-1+\sqrt{5}}{2}~\approx~0,615$

$\sf x'=\dfrac{-1-\sqrt{5}}{2}~\approx~-1,615$

Temos $\sf \dfrac{-1-\sqrt{5}}{2} < x < \dfrac{-1+\sqrt{5}}{2}$

O conjunto solução é:

$\sf S=\Big\{x\in\mathbb{R}~|~ x < \dfrac{-1-\sqrt{21}}{2}~ou~\dfrac{-1-\sqrt{5}}{2} < x < \dfrac{-1+\sqrt{5}}{2}~ou~x > \dfrac{-1+\sqrt{21}}{2}\Big\}$