Question

1) Quais as raízes da equação.
A)
$x(x + 3) = 5x + 15$

B)
$\frac{3y + 1}{2} = \frac{ {y}{2} - 1}{3}$

C)
$(x + 4 {)}^2{} = 9x + 22$

Answer 1

a

x ( x + 3 ) = 5x + 15

x² + 3x = 5x + 15

x² + 3x -5x - 15 = 0

x² - 2x - 15 = 0

delta =( -2)² - [ 4 * 1 * (-15)] =  4 + 60 = 64 ou +-V64 = +-8 ***

x = ( 2 +-8)/2

x1 = 10/2 = 5 **** resposta

x2 = -6/2 = -3 ****resposta

b

( 3y + 1)/2 = ( y² - 1 )/3

multiplique em cruz

2 ( y² - 1)  = 3 ( 3y + 1)

2y² - 2 = 9y + 3

2y² - 9y  -2 - 3 = 0

2y² - 9y - 5 = 0

delta = ( -9)² - [ 4 * 2 * (-5)] =  81 + 40 = 121 ou +-V121 = +-11 ****

x = ( 9 +-11)/4

x1 = 20/4 = 5 **** resposta

x2 = -2/4 = - 1/2 ***resposta

c

( x + 4 )²  = 9x + 22

[ ( x)²  + 2 *  x * 4  + (4)² ]  = 9x + 22

x² + 8x + 16 = 9X+ 22

X²  + 8X + 16  - 9X - 22 = 0

X² - X - 6 = 0

DELTA = (-1)²  - [ 4 * 1 *  ( -6)] =  1 + 24 = v25 OU +-5 ****

X = ( 1 +-5)/2

X1 =  6/2 = 3 ****

X2 = -4/2 = -2 *****

Answer 2

Olá!!!

Resolução!!

a)

x ( x + 3 ) = 5x + 15
x² + 3x = 5x + 15
x² + 3x - 5x - 15 = 0
x² - 2x - 15 = 0

a = 1, b = - 2, c = - 15

∆ = b² - 4ac
∆ = ( - 2 )² - 4 • 1 • ( - 15 )
∆ = 4 + 60
∆ = 64

x = - b ± √∆/2a
x = - ( - 2 ) ± √64/2 • 1
x = 2 ± 8/2
x' = 2 + 8/2 = 10/2 = 5
x" = 2 - 8/2 = - 6/2 = - 3

S = { - 3, 5 }

b)

( 3y + 1 )/2 = ( y² - 1 )/3

Multiplicando a cruzada

2 • ( y² - 1 ) = ( 3y + 1 ) • 3
2y² - 2 = 9y + 3
2y² - 2 - 9y - 3 = 0
2y² - 9y - 3 - 2 = 0
2y² - 9y - 5 = 0

a = 2, b = - 9, c = - 5

∆ = b² - 4ac
∆ = ( - 9 )² - 4 • 2 • ( - 5 )
∆ = 81 + 40
∆ = 121

x = - b ± √∆/2a
x = - ( - 9 ) ± √121/2 • 2
x = 9 ± 11/4
x' = 9 + 11/4 = 20/4 = 5
x" = 9 - 11/4 = - 2/4 ÷ 2 = - 1/2

S = { - 1/2, 5 }

c)

( x + 4 )² = 9x + 22
( x )² + 2 • x • 4 + 4² = 9x + 22
x² + 8x + 16 = 9x + 22
x² + 8x + 16 - 9x - 22 = 0
x² + 8x - 9x - 22 + 16 = 0
x² - x - 6 = 0

a = 1, b = - 1 , c = - 6

∆ = b² - 4ac
∆ = ( - 1 )² - 4 • 1 • ( - 6 )
∆ = 1 + 24
∆ = 25

x = - b ± √∆/2a
x = - ( - 1 ) ± √25/2 • 1
x = 1 ± 5/2
x' = 1 + 5/2 = 6/2 = 3
x" = 1 - 5/2 = - 4/2 = - 2

S = { - 2, 3 }

Espero ter ajudado!!