Question

Dada a função y = 2x² + 5x + 2, determine:

a) Valores de y quando x = 1 ex = -2

b) Valores de x quando y = 0 ey = -1 (Lembre-se: usar Bhaskara)

Answer 1

Resposta:

a)

x = 1 ; y=9

x = -2; y=0

b)

y=0; x= -2 ou x= -1/2

y=-1; x=-1 ou x = -3/2

Explicação passo-a-passo:

$x = 1 \\ y = 2 + 5 + 2 = 9 \\ \\ x = - 2 \\ y = 8 - 10 + 2 = 0$

$quando y = 0 \\ x = \frac{ - 5 + - 3}{4} \\ x1 = \frac{ - 2}{4} = \frac{ - 1}{2} \\ x2 = \frac{ - 5 - 3}{4} = - 2 \\ \\ quando y = - 1 \\ x = \frac{ - 5 + - 1}{4} \\ x1 = \frac{ - 4}{4} = - 1 \\ x = \frac{ - 6}{4} = - \frac{3}{2}$

Answer 2

Explicação passo a passo:

$\sf y = 2x² + 5x + 2$

a) Determinar valores de y

=> Para x = 1

$\sf y = 2(1)^2 + 5(1) + 2$

$\sf y = 2 + 5 + 2$

$\red{\sf y = 9}$

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=> Para x = - 2

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

$\sf y = 2(4) - 10 + 2$

$\sf y = 8 - 10 + 2$

$\red{\sf y = 0}$

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b) Determinar valores de x

=> Para y = 0

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

coeficientes: a = 2, b = 5, c = 2

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

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

$\sf \Delta = 25 - 16$

$\sf \Delta = 9$

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

$\sf x = \dfrac{- (5) \pm \sqrt{9}}{2*(2)}~\Rightarrow~x = \dfrac{- 5 \pm 3}{4}$

$\sf x' = \dfrac{- 5 + 3}{4}~\Rightarrow~x' = - \dfrac{2}{4}~\Rightarrow~x' = \red{-\dfrac{1}{2}}$

$\sf x'' = \dfrac{- 5 - 3}{4}~\Rightarrow~x'' = - \dfrac{8}{4}~\Rightarrow~x'' = \red{- 2}$

o conjunto solução é:

$\boxed{\sf S = \bigg\{- 2~~;~~- \dfrac{1}{2}\bigg\}}$

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=> Para y = - 1

$\sf - 1 = 2x^2 + 5x + 2$

$\sf 0 = 2x^2 + 5x + 2 + 1$

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

coeficientes: a = 2, b = 5, c = 3

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

$\sf \Delta = (5)^2 - 4*(2)*(3)$

$\sf \Delta = 25 - 24$

$\sf \Delta = 1$

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

$\sf x = \dfrac{- (5) \pm \sqrt{1}}{2*(2)}~\Rightarrow~x = \dfrac{- 5 \pm 1}{4}$

$\sf x' = \dfrac{- 5 + 1}{4}~\Rightarrow~x' = - \dfrac{4}{4}~\Rightarrow~x' = \red{-1}$

$\sf x'' = \dfrac{- 5 - 1}{4}~\Rightarrow~x'' = - \dfrac{6}{4}~\Rightarrow~x'' = \red{- \dfrac{3}{2}}$

o conjunto solução é:

$\boxed{\sf S = \bigg\{- \dfrac{3}{2}~~;~~- 1\bigg\}}$