Question

1.x^4-5x^2+4=0

2.125x^3-8=0

3.x^3+3x^2-a=0

Answer 1

1.x^4-5x^2+4=0

x^2 = t i t > 0

t^2 - 5t + 4 = 0

Δ = 5² - 4 * 1 * 4 = 25 - 16 = 9

√Δ = 3

t₁ = (5 - 3) / 2 = 1 > 0

t₂ = (5 + 3) / 2 = 4 > 0

 

x² = t₁                        lub x² = t₂

x² = 1                         lub x² = 4

x = 1  lub x = -1  lub x = 2  lub x = -2

 

2.125x^3-8=0

(5x)³ - 2³ = 0

(5x)³ = 2³

5x = 2

x = 2/5

 

3.x^3+3x^2 - 4 = 0

x^3 - x^2 + 4x^2 - 4 = 0

x^2(x - 1) + 4(x^2 - 1) = 0

x^2(x - 1) + 4(x - 1)(x + 1) = 0

(x - 1)[ x^2 + 4(x + 1) ] = 0

(x - 1)(x^2 + 4x + 4) = 0

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

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

x = 1  lub x = -2  lub x = -2