Question

Rozwiąż równanie - halp!
a) x^2 - 4x = 0
b) x^2 - 8x + 16 = 0
c) 2x^2 + 3 = -7x
d) (1 - 2x)^2 = 2x - x^2 + 1
e) (2x - 1)^2 = 3x + x^2 + 1
zadanie z funkcji kwadratowej

Answer 1

A) x^2 - 4x = 0
Δ = b² - 4ac
Δ = 16 - 4 × 1 × 0
Δ = 16
√Δ = 4
x1= (-b - √Δ) / 2a
x1 = 4 -4 / 2 = 0
x2 = ( -b + √Δ ) / 2a
x2 = (4+4)/2 = 4
b) x^2 - 8x + 16 = 0
Δ = 64 - 4 * 16
Δ = 64 - 64
Δ = 0
x = -b / 2a
x = 8 / 2
x = 4
c) 2x^2 + 3 = -7x
2x^2 + 7x + 3 = 0
Δ = 49 - 8 * 3
Δ = 49 - 24
Δ = 25
√Δ = 5
x1 = (-7 -5) / 4
x1= -12/4
x1 = -3
x2 = -7 + 5 / 4
x2 = -2/4
x2 = -1/2
d) (1 - 2x)^2 = 2x - x^2 + 1
1 - 4x + 4x^2  = 2x - x^2 + 1
5x^2 - 6x = 0
Δ = 36
√Δ = 6
x1 = 6 - 6 / 10
x1 = 0
x2 = 6 + 6 / 10
x2 = 12 /10
x2 = 1,2
e) (2x - 1)^2 = 3x + x^2 + 1
4x^2 - 4x + 1 = 3x + x^2 +1
3x^2 - x = 0
Δ = 1
√Δ = 1
x1 = 1 -1 /6 = 0
x2 = 1+1/ 6 = 2/6 = 1/3