Question

O produto das raizes da equação 4^x-2 - 17ₓ2^x-4 + 1 =0
RESOLVER PASSO A PASSO, principalmente quando for igualar as bases.
USE A IMAGEM PARA MELHOR COMPREENSÃO.

Answer 1

$4^{x-2}~-~17~.~2^{x-4}~+~1~=~0\\\\\\\left(2^2\right)^{x-2}~-~17~.~2^{x-2-2}~+~1~=~0\\\\\\\left(2^{x-2}\right)^2~-~17~.~2^{x-2}~.~2^{-2}~+~1~=~0\\\\\\\left(2^{x-2}\right)^2~-~\frac{17}{4}~.~2^{x-2}~+~1~=~0\\\\\\Substituindo~~2^{x-2}~=~w,~temos:\\\\\\\boxed{w^2~-~\frac{17}{4}w~+ 1~=~0}\\\\\\Vamos~aplicar~Bhaskara\\\\\\\Delta~=~\left(-\frac{17}{4}\right)^2-4.1.1~=~\frac{289}{16}-4~=~\boxed{\frac{225}{4}}$

$w'~=~\dfrac{\frac{17}{4}+\sqrt{\frac{225}{16}}}{2~.~1}~=~\dfrac{\frac{17}{4}+\frac{15}{4}}{2}~=~\frac{32}{4~.~2}~=~\frac{32}{8}~=~\boxed{4}\\\\\\w''~=~\dfrac{\frac{17}{4}-\sqrt{\frac{225}{16}}}{2~.~1}~=~\dfrac{\frac{17}{4}-\frac{15}{4}}{2}~=~\frac{2}{4~.~2}~=~\frac{2}{8}~=~\boxed{\frac{1}{4}}$

Vamos agora voltar a substituição feita:

$\boxed{2^{x-2}~=~w}\\\\\\\\2^{x'-2}~=~w'\\\\\\2^{x'-2}~=~4\\\\\\2\!\!\!\backslash^{x'-2}~=~2\!\!\!\backslash^2\\\\\\x'-2~=~2\\\\\\\boxed{x'~=~4}\\\\\\\\2^{x''-2}~=~w''\\\\\\2^{x''-2}~=~\frac{1}{4}\\\\\\2^{x''-2}~=~\frac{1}{2^2}\\\\\\2\!\!\!\backslash^{x''-2}~=~2\!\!\!\backslash^{-2}\\\\\\x''-2~=~-2\\\\\\\boxed{x''~=~0}$

Sendo assim, o produto das raízes será:

$x'~.~x''~=~4~.~0~=~\boxed{0}~~~\checkmark$

Answer 2

Resposta: alternativa A

4ˣ⁻² - 17. 2ˣ⁻⁴ + 1 = 0

2²⁽ˣ⁻²⁾ - 17 . 2⁽ˣ⁻²⁻²⁾ + 1 = 0

(2ˣ⁻²)² - 17. 2⁽ˣ⁻²⁾ . 2⁻² + 1 = 0

Fazendo 2⁽ˣ⁻²⁾ = y

y² - 17y . $\frac{1}{4}$+1 = 0

MMC = 4

4y² - 17y + 4 = 0

Δ = (-17)² - 4.4.4

Δ = 289 - 64

Δ = 225

y' = [-(-17) + √225]/2.4 ∴ y' = [17 + 15]/8 ∴ y' = 32/8 ∴ y'= 4

y'' = [17-15]/8 ∴ y'' = 2/8 ∴ y'' = 1/2

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Voltando aos 2⁽ˣ⁻²⁾ = y:

2⁽ˣ⁻²⁾ = 4

2⁽ˣ⁻²⁾ = 2²

x-2 = 2

x = 2+2

x = 4

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2⁽ˣ⁻²⁾ = $\frac{1}{4}$

2⁽ˣ⁻²⁾ = 2⁻²

x-2 = -2

x = -2+2

x = 0

As raízes da equação são os valores de x = 4 e 0.

O produto delas é 4.0 = 0