Question

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Answer 1

Resposta:

b)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-6x+9=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-6~e~c=9\\\\\Delta=(b)^{2}-4(a)(c)=(-6)^{2}-4(1)(9)=36-(36)=0\\\\x^{'}=x^{''}=\frac{-(b)\pm\sqrt{\Delta}}{2(a)}=\frac{-(-6)\pm\sqrt{0}}{2(1)}=\frac{6\pm0}{2}=\frac{6}{2}=3\\\\S=\{3,~3\}$

Raiz dupla

c)

x²-49=0

x²=49

x=±√49=±7

S={-7,7}

d)

2x²-4x=0

2x(x-2)=0

2x=0 => x=0

x-2=0=> x=2

S={0,2}