Question

Ajudem-me, por favor! 

Answer 1

$1) \\ \\ 3x^2 + 75 = 0$

Podemos dividir por 3 para facilitar o calculo, não altera os valores das raízes:

$\dfrac{3x^2 + 75 }{3} => x^2 + 25 = 0$

$x^2 + 25 = 0 \\ \\ x^2 = - 25 \\ \\ x = \pm \ \sqrt{-25} \\ \\ x = \pm \ \sqrt{25} * i \\ \\ x' = +\sqrt{25} * i \\ \\ x' = 5 * i \\ \\ x'' = - \sqrt{25} * i \\ \\ x'' = -5 * i$


S = {5*i,  -5 *i}

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2)

$x^2 - 8x + 25 = 0$

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=1, b=−8, c=25

Δ=b²−4ac
Δ=(−8)³−4*1*25
Δ=64−100
Δ= −36

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-(-8) \pm \sqrt{-36}}{2*1} \\ \\ \\ x = \dfrac{8 \pm \sqrt{36}*i}{2} \\ \\ \\ x = \dfrac{8 \pm 6*i}{2} \\ \\ \\ x' = \dfrac{8 + 6*i}{2} \\ \\ \\ x' = 4 + 3 *i \\ \\ \\ x'' = \dfrac{8 - 6*i}{2} \\ \\ \\ x'' = 4 - 3 * i$


S = { 4 + 3 * i,   4 - 3 * i}

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Exercícios

1)

a)

$x^2 + 4 = 0 \\ \\ x^2 = - 4 \\ \\ x = \pm \ \sqrt{-4} \\ \\ x = \pm \sqrt{4} * i \\ \\ x' = 2 *i \\ \\ x'' = -2 * i$


S = {2*i,  -2*i}

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b)

$x^2 + 25 = 0 \\ \\ x^2 = - 25 \\ \\ x = \pm \ \sqrt{-25} \\ \\ x = \pm \ \sqrt{25} * i \\ \\ x' = +\sqrt{25} * i \\ \\ x' = 5 * i \\ \\ x'' = - \sqrt{25} * i \\ \\ x'' = -5 * i$


S = {5*i,  -5*i}

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2)

a)

$x^{2} - 2x + 2 = 0$

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=1, b=−2, c=2

Δ=b²−4ac
Δ=(−2)²−4*1*2
Δ=4−8
Δ=−4

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \dfrac{-(-2) \pm \sqrt{-4}}{2} \\ \\ \\ x = \dfrac{2 \pm \sqrt{4}*i}{2} \\ \\ \\ x = \dfrac{2 \pm 2*i}{2} \\ \\ \\ x' = \dfrac{2 + 2*i}{2} \\ \\ \\ x' = 1 + 1 *i \\ \\ \\ x'' = \dfrac{2 - 2*i}{2} \\ \\ \\ x'' = 1 - 1 * i$


S = {1 + 1 * i,  1 - 1 * i}

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b)

$2x^2 - 6x + 9 = 0$

a=2, b=−6, c=9

Δ=b²−4ac
Δ=(−6)²−4*2*9
Δ=36−72
Δ= −36

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \dfrac{-(-6) \pm \sqrt{-36}}{2*2} \\ \\ \\ x = \dfrac{6 \pm \sqrt{36}*i}{4} \\ \\ \\ x = \dfrac{6 \pm 6*i}{4} \\ \\ \\ x' = \dfrac{6 + 6*i}{4} \\ \\ \\ x' = \dfrac{3 + 3*i}{2} \\ \\ \\ x'' = \dfrac{6 - 6*i}{4} \\ \\ \\ x'' = \dfrac{3 - 3*i}{2}$


S = {$\dfrac{3 + 3*i}{2}, \ \ \dfrac{3 - 3*i}{2}$}

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d)

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

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=1, b=2, c=5

Δ=b²−4ac
Δ=(2)²−4*1*5
Δ=4−20
Δ= −16

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \dfrac{-2 \pm \sqrt{-16}}{2*1} \\ \\ \\ x = \dfrac{-2 \pm \sqrt{16}*i}{2} \\ \\ \\ x = \dfrac{-2 \pm 4*i}{2} \\ \\ \\ x' = \dfrac{-2 + 4*i}{2} \\ \\ \\ x' = -1 + 2 * i \\ \\ \\ x'' = \dfrac{-2 - 4*i}{2} \\ \\ \\ x'' = -1 - 2 * i$


S = {-1 + 2 * i,   -1 - 2 * i}

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e)

$3t^2 + t + 1 = 0$

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=3, b=1, c=1
Δ=b²−4ac
Δ=(1)²−4*3*1
Δ=1−12
Δ= −11

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-1 \pm \sqrt{-11}}{2*3} \\ \\ \\ x = \dfrac{-1 \pm \sqrt{11}*i}{6} \\ \\ \\ x' = \dfrac{-1 + \sqrt{11}*i}{6} \\ \\ \\ x'' = \dfrac{-1 - \sqrt{11}*i}{6}$


S = {$\dfrac{-1 + \sqrt{11}*i}{6} , \ \ \dfrac{-1 - \sqrt{11}*i}{6}$}