Question

$\sqrt{1 - \times } = x + 5 \: vom \\ 1 - x \geqslant 0$
Me ajudem pfvr​

Answer 1

Resposta:

x= -3

Explicação passo-a-passo:

$\sqrt{1-x} = x + 5\\(\sqrt{1-x})^{2} =(x + 5)^{2}\\1-x=x^{2}+10x+25\\x^{2}+11x+24=0\\\\Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+11x+24=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=11~e~c=24\\\\\Delta=(b)^{2}-4(a)(c)=(11)^{2}-4(1)(24)=121-(96)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(11)-\sqrt{25}}{2(1)}=\frac{-11-5}{2}=\frac{-16}{2}=-8\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(11)+\sqrt{25}}{2(1)}=\frac{-11+5}{2}=\frac{-6}{2}=-3\\\\S=\{-8,~-3\}$

Fazendo a prova:

$Para~x=-8\\\sqrt{1-x} = x + 5\\\sqrt{1-(-8)} = -8 + 5\\\sqrt{9} = -3\\3=-3~(F)\\\\Para~x=-3\\\sqrt{1-x} = x + 5\\\sqrt{1-(-3)} = (-3) + 5\\\sqrt{4} = 2\\2=2~(V)$