Question

Escreva as equações na forma geral e resolva. ​Utilizando a fórmula de bhaskara.​

Answer 1

d)

$4 {x}^{2} + 7x + 3 = 2 {x}^{2} + 2x \\ 2 {x}^{2} + 5x + 3 = 0 \\ Δ = 25 - 4 \times 2 \times 3 \\ Δ = 25 - 24 = 1 \\ x = \frac{ - 5 \frac{ + }{} \sqrt{1} }{4} \\ x = - 1 \: ou \: x = - \frac{3}{2}$

e)

$x(x -2 ) = 2(x + 6) \\ {x}^{2} - 2x = x + 12 \\ {x}^{2} - 3x - 12 = 0 \\ Δ = 9 - 4 \times 1 \times ( - 12) \\ Δ = 9 + 48 = 57 \\ x = \frac{3 \frac{ + }{} \sqrt{57} }{2}$

f)

$x(2x - 1) + 6 = 4(x + 1) \\ 2 {x}^{2} - x + 6 = 4x + 4 \\ 2 {x}^{2} - 5x + 2 = 0 \\ Δ = 25 - 4 \times 2 \times 2 \\ Δ = 25 - 16 = 9 \\ x = \frac{ 5 \frac{ + }{ } \sqrt{9} }{4} \\ x = 2 \: ou \: x = \frac{1}{2}$

g)

$(x - 1)(x - 2) = 6 \\ {x}^{2} - 2x - x + 2 = 6 \\ {x}^{2} - 3x - 4 = 0 \\ Δ = 9 - 4 \times 1 \times ( - 4) \\ Δ = 9 + 16 = 25 \\ x = \frac{3 \frac{ + }{} \sqrt{25} }{2} \\ x = 4 \: ou \: x = - 1$

h)

$(2x - 3)(x - 8) = 34 \\ 2 {x}^{2} - 16x - 3x + 24 = 34 \\ 2 {x}^{2} - 16x - 10 = 0 \: \div (2) \\ {x}^{2} - 8x - 5 = 0 \\ Δ = 64 - 4 \times 1 \times ( - 5) \\ Δ = 64 + 20 = 84 \\ x = \frac{ 8 \frac{ + }{} \sqrt{84} }{2} \\ x = 4 \frac{ + }{} \sqrt{21}$

Answer 2

Olá...

d)

4x² + 7x + 3 = 2x² + 2x

4x² + 7x + 3 - 2x² - 2x = 0

2x² + 5x + 3 = 0

2x² + 3x +2x + 3 = 0

x(2x + 3) + 2x + 3 = 0

(2x + 3).(x + 1) = 0

2x + 3 = 0

2x = - 3

x' = - 3/2

x + 1 = 0

x" = - 1

S = {- 3/2; - 1}

e)

x(x - 2) = 2(x + 6)

x² - 2x = 2x + 12

x² - 2x - 2x - 12 = 0

x² - 4x - 12 = 0

x(x + 2) - 6(x + 2) = 0

(x + 2).(x - 6) = 0

x + 2 = 0

x' = - 2

x - 6 = 0

x" = 6

S = {- 2; 6}

f)

x(2x - 1) + 6 = 4(x + 1)

2x² - x + 6 = 4x + 4

2x² - x + 6 - 4x - 4 = 0

2x² - 5x + 2 = 0

2x² - x - 4x + 2 = 0

x(2x - 1) - 2(2x - 1) = 0

(2x - 1).(x - 2) = 0

2x - 1 = 0

2x = 1

x' = 1/2

x - 2 = 0

x" = 2

S = {1/2; 2}

g)

(x - 1).(x - 2) = 6

x² - 2x - x + 2 = 6

x² - 3x + 2 = 6

x² - 3x + 2 - 6 = 0

x² + x - 4x - 4 = 0

x(x + 1) - 4(x + 1) = 0

(x + 1).(x - 4) = 0

x + 1 = 0

x' = - 1

x - 4 = 0

x" = 4

S = {- 1; 4}

h)

(2x - 3).(x - 8) = 34

2x² - 16x - 3x + 24 = 34

2x² - 19x + 24 - 34 = 0

2x² + x - 20x - 10 = 0

x(2x + 1) - 10(2x + 1) = 0

(2x + 1).(x - 10) = 0

2x + 1 = 0

2x = - 1

x' = - 1/2

x - 10 = 0

x" = 10

S = {- 1/2; 10}

Espero ter ajudado.