помогите срочно [tex]x {}^{2} + |x + 4| = 4[/tex]. Created by fsaken. algebra-ru

tex]...

Universalka

$x^{2}+|x+4|=4\\\\1)x-4-neyd\\\\x_{2}=\dfrac{1+\sqrt{33} }{2} >-4-neyd$

$2) \ x\geq- 4\\\\x^{2} +x+4=4\\\\x^{2}+x=0\\\\x(x+1)=0\\\\x_{1}=0\\\\x+1=0\\\\x_{2} =-1\\\\Otvet: \boxed{-1 \ ; \ 0}$

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dobra888

Ответ: x₁ = - 1 ; x₂ = 0 .

Объяснение:

1 . x² + | x + 4 | = 4 ;

x + 4 = 0 ; x = - 4 ;

a ) x ≤ - 4 ; x² - ( x + 4 ) = 4 ; x² - x - 8 = 0 ; D = 33 > 0 ;

x₁ = ( 1 - √ 33 )/2 ; x₂ = ( 1 + √ 33 )/2 ; x₁ i x₂ не належать проміжку

б ) x > - 4 ; x² + ( x + 4 ) = 4 ; x² + x = 0 ; x₁ = - 1 ; x₂ = 0 .

обидва ці корені належать другому проміжку .

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