Uma outra forma interessante de resolver. Completando o quadrado com os 2 primeiros termos: a million/x-4/raiz(x)+3=0 [a million/x-4/raiz(x)+4]-a million=0 [a million/raiz(x)-2]²=a million a million/raiz(x)-2=+-a million a) a million/raiz(x)-2=a million a million/raiz(x)=3 raiz(x)=a million/3 x=a million/9 b) a million/raiz(x)-2=-a million a million/raiz(x)=a million raiz(x)=a million x=a million resposta....a million/9 e a million
Comments
3^(4√x) - 4*3^(√4x) + 3 = 0
3^(4√x) - 4*3^(2√x) + 3 = 0
[3^(2√x)]² - 4*3^(2√x) + 3 = 0 -> façamos a = 3^(2√x)
a² - 4a + 3 = 0
a' = 1 e a'' = 3
Para a = 1:
1 = 3^(2√x) = 3º
2√x = 0
x = 0
Para a = 3:
3 = 3^(2√x)= 3¹
2√x = 1
√x = 1/2
x = 1/4
Então,
16(p+q) =
= 16(0 + 1/4)
= 16.(1/4)
= 16/4
= 4
Alternativa b)
Uma outra forma interessante de resolver. Completando o quadrado com os 2 primeiros termos: a million/x-4/raiz(x)+3=0 [a million/x-4/raiz(x)+4]-a million=0 [a million/raiz(x)-2]²=a million a million/raiz(x)-2=+-a million a) a million/raiz(x)-2=a million a million/raiz(x)=3 raiz(x)=a million/3 x=a million/9 b) a million/raiz(x)-2=-a million a million/raiz(x)=a million raiz(x)=a million x=a million resposta....a million/9 e a million