Question

$\sqrt{x + 2} = \sqrt{3x - 5} - 1$
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Answer 1

Resposta:

x=7

Explicação passo-a-passo:

$\sqrt{x+2}=\sqrt{3x-5} -1\\(\sqrt{x+2})^{2}=(\sqrt{3x-5} -1)^{2}\\x+2=3x-5-2\sqrt{3x-5}+1\\2\sqrt{3x-5}=2x-6\\\sqrt{3x-5}=x-3\\(\sqrt{3x-5})^{2}=(x-3)^{2}\\3x-5=x^{2}-6x+9\\x^{2}-9x+14=0$

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-9x+14=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-9~e~c=14\\\\\Delta=(b)^{2}-4(a)(c)=(-9)^{2}-4(1)(14)=81-(56)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-9)-\sqrt{25}}{2(1)}=\frac{9-5}{2}=\frac{4}{2}=2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-9)+\sqrt{25}}{2(1)}=\frac{9+5}{2}=\frac{14}{2}=7$

Para x=2

$\sqrt{2+2}=\sqrt{3.2-5} -1\\\sqrt{4} =\sqrt{6-5} -1\\2=\sqrt{1}-1\\2=1-1\\2=0~(Falso)$

Para x=7

$\sqrt{7+2}=\sqrt{3.7-5} -1\\\sqrt{9}=\sqrt{21-5} -1\\3=\sqrt{16}-1\\3=4-1\\3=3~(Verdadeiro)$