Question

Usando a fórmula de Bhaskara, resolva, em R, a equação dada:
respondam pfvvvv <3

Answer 1

Resposta:

Explicação passo-a-passo:

(x²+3)/10-(5x+3)/15=(3-2x)/6

mmc(6,10,15)

6, 10, 15 | 2

3, 5, 15 | 3

1, 5, 5 | 5

1, 1, 1 | 1

mmc(6,10,15) = 2 . 3 . 5 = 30

3(x²+3)/30-2(5x+3)/30=5(3-2x)/30

3(x²+3)-2(5x+3)=5(3-2x)

3x²+9-10x-6=15-10x

3x²-10x+10x+3-15=0

3x²-12=0 ÷(3)

x²-4=0

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-0x-4=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=0~e~c=-4\\\\\Delta=(b)^{2}-4(a)(c)=(0)^{2}-4(1)(-4)=0-(-16)=16\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(0)-\sqrt{16}}{2(1)}=\frac{0-4}{2}=\frac{-4}{2}=-2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(0)+\sqrt{16}}{2(1)}=\frac{0+4}{2}=\frac{4}{2}=2\\\\S=\{-2,~2\}$