Question

Dado a função y = x² - 6x + 5 determinar: a) Valores de y para x = 3 ex = -2 (SUBSTITUIR X)

b) Valores de x para y = 0 ey = -3 (USAR BHASKARA)

Answer 1

Explicação passo-a-passo:

$\sf y = x² - 6x + 5$

a) Determinar valores de y

=> Para x = 3

$\sf y = (3)^2 - 6(3) + 5$

$\sf y = 9 - 18 + 5$

$\red{\sf y = - 4}$

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=> Para x = - 2

$\sf y = (-2)^2 - 6(-2) + 5$

$\sf y = 4 + 12 + 5$

$\red{\sf y = 21}$

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b) Determinar valores de x

=> Para y = 0

$\sf 0 = x^2 - 6x + 5$

coeficientes: a = 1, b = - 6, c = 5

$\sf \Delta = b^2 - 4ac$

$\sf \Delta = (-6)^2 - 4*(1)*(5)$

$\sf \Delta = 36 - 20$

$\sf \Delta = 16$

$\sf x = \dfrac{- b \pm \sqrt{\Delta}}{2a}$

$\sf x = \dfrac{- (-6) \pm \sqrt{16}}{2*(1)}~\Rightarrow~x = \dfrac{6 \pm 4}{2}$

$\sf x' = \dfrac{6 + 4}{2}~\Rightarrow~x' = \dfrac{10}{2}~\Rightarrow~x' = \red{5}$

$\sf x'' = \dfrac{6 - 4}{2}~\Rightarrow~x'' = \dfrac{2}{2}~\Rightarrow~x'' = \red{1}$

o conjunto solução é:

$\boxed{\sf S = \left\{1~~;~~5\right\}}$

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=> Para y = - 3

$\sf - 3 = x^2 - 6x + 5$

$\sf 0 = x^2 - 6x + 5 + 3$

$\sf 0 = x^2 - 6x + 8$

coeficientes: a = 1, b = - 6, c = 8

$\sf \Delta = b^2 - 4ac$

$\sf \Delta = (-6)^2 - 4*(1)*(8)$

$\sf \Delta = 36 - 32$

$\sf \Delta = 4$

$\sf x = \dfrac{- b \pm \sqrt{\Delta}}{2a}$

$\sf x = \dfrac{- (-6) \pm \sqrt{4}}{2*(1)}~\Rightarrow~x = \dfrac{6 \pm 2}{2}$

$\sf x' = \dfrac{6 + 2}{2}~\Rightarrow~x' = \dfrac{8}{2}~\Rightarrow~x' = \red{4}$

$\sf x'' = \dfrac{6 - 2}{2}~\Rightarrow~x'' = \dfrac{4}{2}~\Rightarrow~x'' = \red{2}$

o conjunto solução é:

$\boxed{\sf S = \left\{2~~;~~4\right\}}$