Matemática, perguntado por Emmacloud, 7 meses atrás

Determine o valor de x , me ajudem pfv <3

Anexos:

Soluções para a tarefa

Respondido por MateusD3
1

b) S = { -3, -20}

c) S = { 6, -4 }

d) S = {3, -4}

e) S = {7, 6}

f) S = {8, -3}

g) S = {-1, 2}

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b) x² + 13x + 30 = 0

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

a = 1

b = 13

c = 30

Δ = b² - 4.a.c

Δ = 13² - 4 . 1 . 30

Δ = 169 - 120

Δ = 49

\sf  x= \dfrac{-13 \pm \sqrt{49} }{2}

{x_1} = \dfrac{ -13 + 7}{2}

{x_1} = \dfrac{ -6}{2}

\boxed{x_1 = -3}

\boxed{x_2 = -20}

c) x² - 2x - 24 = 0

a = 1

b = -2

c = -24

Δ = (-2)² - 4 . 1 . -24

Δ = 4 - 4 . -24

Δ = 100

\sqrt{100} = 10

x' =\dfrac{2 + 10}{2(1)}

x' = 6

x'' = \dfrac{2 - 10}{2(1)}

x'' = -4

d) x² + 1x - 12 = 0

a = 1

b = 1

c = -12

Δ = 1 - 4 . -12

Δ = 1 + 48

Δ = 49

\sqrt{49} = 7

x' = \dfrac{-1 + 7}{2(1)}

x'' = \dfrac{-1 - 7}{2(1)}

x' = 3

x'' = -4

e) x² - 13x + 42 = 0

a = 1

b = -13

c = 42

Δ = (-13)² - 4 * 1 * 42

Δ = 169 - 168

Δ = 1

\sqrt{1} = 1

x' = \dfrac{13 + 1}{2(1)} \\\\\\\\\ x'' = \dfrac{13 - 1}{2(1)}

x' = 7 \\\\ x'' = 6

f) 2x² - 10x - 48 = 0

a = 2

b = -10

c = -48

Δ = (-10)² - 4 * 2 * -48

Δ = 100 + 384

Δ = 484

\sqrt{484} = 22

x' = \dfrac{10 + 22}{4} \\\\\\ x'' = \dfrac{10 - 22}{4}

x' = 8 \\\\\ x'' = -3

g) -5x² + 5x + 10 = 0

a = - 5

b = 5

c = 10

Δ = 5² - 4 * -5 *  10

Δ = 25 + 200

Δ = 225

\sqrt{225} = 15

x' = \dfrac{-5 + 15}{-10} \\\\\\\ x'' = \dfrac{-5 - 15}{-10}

x' = -1 \\\\ x'' = 2


Emmacloud: muito obrigada <3
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