Odpowiedź:
a)
- 2x² - 12x + 9 = 0
a = - 2 , b = - 12 , c = 9
Δ = b² - 4ac = (- 12)² - 4 * (- 2) * 9 = 144 + 72 = 216
√Δ = √216 = √(36 * 6) = 6√6
x₁ = (- b - √Δ)/2a = (12 - 6√6)/(- 4) = 6(2 - √6)/( - 4) = - 3(2 - √6)/2
x₂ = (- b + √Δ)/2a = (12 + 6√6)/(- 4) = 6(2 + √6)/(- 4) = - 3(2 + √6)/2
b)
(5 - 3x)² = 25 - 9x²
25 - 30x + 9x² = 25 - 9x²
9x² + 9x² - 30x + 25 - 25 = 0
18x² - 30x = 0
6x(3x - 5) = 0
6x = 0 ∨ 3x - 5 = 0
x = 0 ∨ 3x = 5
x = 0 ∨ x = 5/3
x = 0 ∨ x = 1 2/3
c)
1/3x² - x - 2 = 0 | * 3
x² - 3x - 6 = 0
a = 1 , b = - 3 , c = - 6
Δ = b² - 4ac = (- 3)² - 4 * 1 * (- 6) = 9 + 36 = 45
√Δ = √45 = √(9 * 5) = 3√5
x₁ = (- b - √Δ)/2a = (3 - 3√5)/2 = 3(1 - √5)/2
x₂ = (- b + √Δ)/2a = (3 + 3√5)/2 = 3(1 + √5)/2
d)
(x - 3)² - 9 = 0
x² - 6x + 9 - 9 = 0
x² - 6x = 0
x(x - 6) = 0
x = 0 ∨ x - 6 = 0
x = 0 ∨ x = 6
e)
2x² = 5
2x² - 5 = 0
a = 2 , b = 0 , c = - 5
Δ = b² - 4ac = 0² - 4 * 2 * (- 5) = 0 + 40 = 40
√Δ = √40 = √(4 * 10) = 2√10
x₁ = (- b - √Δ)/2a = (0 - 2√10)/4 = - 2√10/4 = - √10/2
x₂ = (- b + √Δ)/2a = (0 + 2√10)/4 = 2√10/4 = √10/2
f)
2x² = - 5
2x² + 5 = 0
Ponieważ x² ≥ 0 dla x ∈ R więc 2x² + 5 > 0 dla x ∈ R
Nie ma takiego x w zbiorze liczb rzeczywistych spełniających to rownanie
x ∈ ∅ (zbiór pusty)
g)
x² - 6x + 8 = 0
a = 1 , b = - 6 , c = 8
Δ = b² - 4ac = (- 6)² - 4 * 1 * 8 = 36 - 32 = 4
√Δ = √4 = 2
x₁ = (- b - √Δ)/2a = (6 - 2)/2 = 4/2 = 3
x₂ = (- b + √Δ)/2a = (6 + 2)/2 = 8/2 = 4
3(x - 1/3)² - 3 = 0
3(x² - 2x/3 + 1/9) - 3 = 0
3x² - 2x + 1/3 - 3 = 0 | * 3
9x² - 6x + 1 - 9 = 0
9x² - 6x - 8 = 0
a = 9 , b = - 6 , c = - 8
Δ = b² - 4ac = (- 6)² - 4 * 9 * (- 8) = 36 + 288 = 324
√Δ = √324 = 18
x₁ = (- b - √Δ)/2a = (6 - 18)/18 = - 12/18 = - 2/3
x₂ = (- b + √Δ)/2a = (6 + 18)/18 = 24/18 = 4/3 = 1 1/3
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Odpowiedź:
a)
- 2x² - 12x + 9 = 0
a = - 2 , b = - 12 , c = 9
Δ = b² - 4ac = (- 12)² - 4 * (- 2) * 9 = 144 + 72 = 216
√Δ = √216 = √(36 * 6) = 6√6
x₁ = (- b - √Δ)/2a = (12 - 6√6)/(- 4) = 6(2 - √6)/( - 4) = - 3(2 - √6)/2
x₂ = (- b + √Δ)/2a = (12 + 6√6)/(- 4) = 6(2 + √6)/(- 4) = - 3(2 + √6)/2
b)
(5 - 3x)² = 25 - 9x²
25 - 30x + 9x² = 25 - 9x²
9x² + 9x² - 30x + 25 - 25 = 0
18x² - 30x = 0
6x(3x - 5) = 0
6x = 0 ∨ 3x - 5 = 0
x = 0 ∨ 3x = 5
x = 0 ∨ x = 5/3
x = 0 ∨ x = 1 2/3
c)
1/3x² - x - 2 = 0 | * 3
x² - 3x - 6 = 0
a = 1 , b = - 3 , c = - 6
Δ = b² - 4ac = (- 3)² - 4 * 1 * (- 6) = 9 + 36 = 45
√Δ = √45 = √(9 * 5) = 3√5
x₁ = (- b - √Δ)/2a = (3 - 3√5)/2 = 3(1 - √5)/2
x₂ = (- b + √Δ)/2a = (3 + 3√5)/2 = 3(1 + √5)/2
d)
(x - 3)² - 9 = 0
x² - 6x + 9 - 9 = 0
x² - 6x = 0
x(x - 6) = 0
x = 0 ∨ x - 6 = 0
x = 0 ∨ x = 6
e)
2x² = 5
2x² - 5 = 0
a = 2 , b = 0 , c = - 5
Δ = b² - 4ac = 0² - 4 * 2 * (- 5) = 0 + 40 = 40
√Δ = √40 = √(4 * 10) = 2√10
x₁ = (- b - √Δ)/2a = (0 - 2√10)/4 = - 2√10/4 = - √10/2
x₂ = (- b + √Δ)/2a = (0 + 2√10)/4 = 2√10/4 = √10/2
f)
2x² = - 5
2x² + 5 = 0
Ponieważ x² ≥ 0 dla x ∈ R więc 2x² + 5 > 0 dla x ∈ R
Nie ma takiego x w zbiorze liczb rzeczywistych spełniających to rownanie
x ∈ ∅ (zbiór pusty)
g)
x² - 6x + 8 = 0
a = 1 , b = - 6 , c = 8
Δ = b² - 4ac = (- 6)² - 4 * 1 * 8 = 36 - 32 = 4
√Δ = √4 = 2
x₁ = (- b - √Δ)/2a = (6 - 2)/2 = 4/2 = 3
x₂ = (- b + √Δ)/2a = (6 + 2)/2 = 8/2 = 4
3(x - 1/3)² - 3 = 0
3(x² - 2x/3 + 1/9) - 3 = 0
3x² - 2x + 1/3 - 3 = 0 | * 3
9x² - 6x + 1 - 9 = 0
9x² - 6x - 8 = 0
a = 9 , b = - 6 , c = - 8
Δ = b² - 4ac = (- 6)² - 4 * 9 * (- 8) = 36 + 288 = 324
√Δ = √324 = 18
x₁ = (- b - √Δ)/2a = (6 - 18)/18 = - 12/18 = - 2/3
x₂ = (- b + √Δ)/2a = (6 + 18)/18 = 24/18 = 4/3 = 1 1/3