Question

resolva a equação modular:​

Answer 1

$\mathsf{|x|}=\begin{cases}\mathsf{x,se~x\ge0}\\\mathsf{-x,se~x\textless0}\end{cases}$

$\mathsf{x^2-3\cdot|x|-4=0}$

$\mathsf{se~x\ge0:}$

$\mathsf{x^2-3x-4=0}\\\mathsf{a=1~~b=-3~~c=-4}\\\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=(-3)^2-4\cdot1\cdot(-4)}\\\mathsf{\Delta=9+16}\\\mathsf{\Delta=25}\\\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\mathsf{x=\dfrac{-(-3)\pm\sqrt{25}}{2\cdot1}}\\\mathsf{x=\dfrac{3\pm5}{2}}\begin{cases}\mathsf{x_{1}=\dfrac{3+5}{2}=\dfrac{8}{2}=4}\\\mathsf{x_{2}=\dfrac{3-5}{2}=-\dfrac{2}{2}=-1}\end{cases}$

$\mathsf{como~x\ge0\implies~x=4}$

$\mathsf{se~x\textless0:}$

$\mathsf{x^2-3\cdot(-x)-4=0\implies~x^2+3x-4=0}\\\mathsf{a=1~~b=3~~c=-4}\\\mathsf{\Delta=3^2-4\cdot1\cdot(-4)}\\\mathsf{\Delta=9+16}\\\mathsf{\Delta=25}\\\mathsf{x=\dfrac{-3\pm\sqrt{25}}{2\cdot1}}\\\mathsf{x=\dfrac{-3\pm5}{2}}\begin{cases}\mathsf{x_{1}=\dfrac{-3+5}{2}=\dfrac{2}{2}=1}\\\mathsf{x_{2}=\dfrac{-3-5}{2}=-\dfrac{8}{2}=-4}\end{cases}$

$\mathsf{como~x\textless0\implies~x=-4}$

Portanto o conjunto solução é

$\huge\boxed{\boxed{\boxed{\boxed{\mathsf{s=\{-4,4\}}}}}}$

Answer 2

Resposta:

Explicação passo-a-passo:

Temos a equação

x² - 3.|x| - 4 = 0

Temos |x| = x, se x ≥ 0, então

x² - 3x - 4 = 0

Δ = (-3)² - 4.1.(-4)

Δ = 9 + 16

Δ = 25

x = (3 ± √25)/2.1

x' = (3 + 5)/2 = 8/2 = 4

x" = (3 - 5)/2 = -2/2 = -1

Mas, a condição exige x ≥ 0, logo x = 4

Temos |x| = -x, se x < 0, então

x² - 3(-x) - 4 = 0 => x² + 3x - 4 = 0

Δ = 3² - 4.1.(-4)

Δ = 9 + 16

Δ = 25

x = (-3 ± √25)/2.1

x' = (-3 + 5)/2 = 2/2 = 1

x" = (-3 - 5)/2 = -8/2 = -4

Mas, a condição exige x < 0, logo x = -4

Portanto, S = {-4, 4}