Question

Resolve as atividades de equações do 2º grau sem o termo b.

a)4X2 +18 = 2X2 +43
b)2X2 – 16 = X2
c)X2 +50= 25
d)-25X2 =9= 0
e)4X2 – 27 = X2
f)5Y2= 25

Answer 1

Resposta:

a) x= ±$\frac{5\sqrt{2} }{2}$    b) x= ±4   c)x= ±5 d) inconclusiva   e )x=±3   f)y=±√5

Explicação passo-a-passo:

a)4X² +18 = 2X² +43

4X² +18 - 2X² -43

2x²-25=0

2x²= 25

x²=25÷2

x= ±$\frac{5\sqrt{2} }{2}$

b) 2x²-16=x²

2x²-x²=16

x²=16

x=±√16

x= ±4

c)-x²= +50-25

-x²=25

-x=±√25

-x=±5

x= ±5

d)-25X2 =9= 0 (INCONCLUSIVA)

e)4X²-27= x²

3x² =27

x²= 27÷3

x²=9

x=±√9

x=±3

f)5y²=25

y²= $\frac{25}{5}$

y²=5

y=±√5

Answer 2

Resposta:

$a)~~4x^{2} +18=2x^{2} +43\\ \\ 4x^{2} -2x^{2} +18-43=0\\ \\ 2x^{2} -25=0~~~~coeficientes:a=2,~b=0,~c=-25\\ \\ x=\sqrt{\frac{-c}{a} } \\ \\ x=\sqrt{\frac{-(-25)}{2} } \\ \\ x=\sqrt{\frac{25}{2} } =\frac{\sqrt{25} }{\sqrt{2} } =\frac{5}{\sqrt{2} } \\ \\ x'=-\frac{5}{\sqrt{2} } \\ \\ x''=\frac{5}{\sqrt{2} }$

$b)~~2x^{2} -16=x^{2} \\ \\ 2x^{2} -x^{2} -16=0\\ \\ x^{2} -16=0~~~~coeficientes:a=1,~b=0,~c=-16\\ \\ x=\sqrt{\frac{-c}{a} } \\ \\ x=\sqrt{\frac{-(-16)}{1} } \\ \\ x=\sqrt{16} \\ \\ x'=-4\\ \\ x''=4$

$c)~~x^{2} +50=25\\ \\ x^{2} +50-25=0\\ \\ x^{2} +25=0~~~~coeficientes:a=1,~b=0,~c=25\\ \\ x=\sqrt{\frac{-c}{a} } \\ \\ x=\sqrt{\frac{-25}{1} } \\ \\ x=\sqrt{-25} ~~~~nao~pertence~ao~conjunto~dos~reais$

$d)~~-25x^{2} +9=0~~coeficioentes:a=-25,~b=0,~c=9\\ \\ x=\sqrt{\frac{-c}{a} } \\ \\ x=\sqrt{\frac{-9}{-25} } =\sqrt{\frac{+9}{+25} } =\frac{\sqrt{9} }{\sqrt{25} } =\frac{3}{5} \\ \\ x'=-\frac{3}{5} \\ \\ x''=\frac{3}{5}$

$e)~~4x^{2} -27=x^{2} \\ \\ 4x^{2} -x^{2} -27=0\\ \\ 3x^{2} -27=0~~~~coeficientes:a=3,~b=0,~c=-27\\ \\ x=\sqrt{\frac{-c}{a} } \\ \\ x=\sqrt{\frac{-(-27)}{3} } \\ \\ x=\sqrt{\frac{27}{3} } =\sqrt{9} =3\\ \\ x'=-3\\ \\ x''=3$

$f)~~5y^{2} =25\\ \\ 5y^{2} -25=0~~~~coeficientes:~a=5,~b=0,~c=-25\\ \\ x=\sqrt{\frac{-c}{a} } \\ \\ x=\sqrt{\frac{-(-25)}{5} } =\sqrt{\frac{25}{5} } \\ \\ x'=-\sqrt{5} \\ \\ x''=\sqrt{5}$