Question

Ajudaa!!

a) 4x² + 9 = 12x
b) x² = x + 12
c) 2x² = -12x - 18
d) x² + 9 = 4x
e) 25x² = 20x - 4
f) 2x= 15- x²
g) 4x² - x + 1 = x + 3x²
H) 3x² + 5x =-x - 9 + 2x²
i) 4 + x ( x - 4) = x
j) x ( x + 3) - 40 =0

Answer 1

a) 4x² + 9 = 12x
4x² + 9 - 12x = 0
4x² - 12x + 9 = 0
a = 4
b = -12
c = 9
Δ = b² - 4ac
Δ = (-12)² - 4(4)(9)
Δ = 144 - 144
Δ = 0
se
Δ = 0
então
x = - b/2a
x = -(-12)/2(4)
x = + 12/8
x = 3/2

b) x² = x + 12
x² - x - 12 = 0
a = 1
b = -1
c = -12
Δ = b² - 4ac
Δ = (-1)² - 4(1)(-12)
Δ = + 1+ 48
Δ = 49----------------------------√49 = 7
se
Δ > 0
então(baskara)
x = - b - + √Δ/2a
x' = -(-1) - √49/2(1)
x' = + 1 - 7/2
x' = -6/2
x' = - 3
x" = -(-1) + √49/2(1)
x" = + 1 + 7/2
x" = 8/2
x" = 4

c) 2x² = -12x - 18
2x² + 12x + 18 = 0
a = 2
b = 12
c = 18
Δ = b² - 4ac
Δ =(12)² - 4(2)(18)
Δ = 144 - 144
Δ = 0
se
Δ = 0
então
x = - b/2a
x = -12/2(2)
x= -12/4
x = - 3

d) x² + 9 = 4x
x² + 9 - 4x = 0
x² -4x + 9 = 0
a = 1
b = - 4
c = 9
Δ = b² -4ac
Δ =(-4)² - 4(1)(9)
Δ = 16 - 36
Δ = - 20
se
Δ < 0  não tem ZEROS reais

e) 25x² = 20x - 4
25x² - 20x + 4 = 0
a = 25
b = -20
c = 4
Δ = b² - 4ac
Δ = (-20)² - 4(25)(4)
Δ = 400 - 400
Δ = 0
se
Δ = 0
então
x = -b/2a
x = -(-20)/2(25)
x = + 20/50
x = 2/5

f) 2x= 15- x²
2x - 15 + x² = 0
x² + 2x - 15 = 0
a = 1
b = 2
c = -15
Δ = b² - 4ac
Δ = (2)²-4(1)(-15)
Δ = 4+ 60
Δ = 64 --------------------------√64 = 8
se
Δ > 0
então(baskara)
x = -b - + √Δ/2A
X' = -2 - √64/2(1)
x' = - 2 -8/2
x' = -10/2
x' = - 5
x" = -2 + √64/2(1)
x" = -2+ 8/2
x" = 6/2
x" = 3

g) 4x² - x + 1 = x + 3x²
4x² - x + 1 - x - 3x² = 0
x² - 2x + 1 = 0
a = 1
b = - 2
c = 1
Δ = b² - 4ac
Δ = (-2)² -4(1)(1)
Δ = 4 - 4
Δ = 0
se
Δ = 0
então
x = -b/2a
x = -(-2)/2(1)
x = +2/2
x = 1


H) 3x² + 5x =-x - 9 + 2x²
3x² + 5x + x + 9 - 2x² = 0
x² + 6x + 9 = 0
a = 1
b = 6
c = 9
Δ = b² - 4ac
Δ = (6)² - 4(1)(9)
Δ = 36 - 36
Δ = 0
se
Δ = 0
então
x = - b/2a
x = -6/2(1)
x = -6/2
x = -3
 
i) 4 + x ( x - 4) = x
4 + x² - 4x = x
4 + x² - 4x - x = 0
x² - 5x + 4 = 0
a = 1
b = - 5
c = 4
Δ = b² - 4ac
Δ = (-5)² - 4(1)(4)
Δ = 25 - 16
Δ = 9 -----------------------√9 = 3
se
Δ > 0
então (baskara)
x = -b - + √Δ/2a
x = -(-5) -√9/2(1)
x = + 5 - 3/2
x = 2/2
x = 1

x" = -(-5) + √9/2(1)
x" = + 5 + 3/2
x" = 8/2
x" = 4

j) x ( x + 3) - 40 =0
x² + 3x - 40 = 0
a = 1
b = 3
c = - 40
Δ = b² - 4ac
Δ = (3)² - 4(1)(-40)
Δ = 9 + 160
Δ = 169-------------------------√169 = 13
se
Δ > 0
então (baskara)
x= - b - + √Δ/2a
x' = -3 - √169/2(1)
x' = -3 - 13/2
x' = -16/2
x' = - 8
x" = -3 + √169/2(1)
x" = -3 + 13/2
x" = 10/2
x" = 5