Question

rozwiąż równanie 4x^4+8x^3+12x^2=0

Answer 1

8x4 + -4x3 + 12x2 = 0

12x2 + -4x3 + 8x4 = 0

12x2 + -4x3 + 8x4 = 0

 

4x2(3 + -1x + 2x2) = 0

 

x2 = 0

x2 = 0

 

x2 = 0

x = {0}

3 + -1x + 2x2 = 0

3 + -1x + 2x2 = 0

 

1.5 + -0.5x + x2 = 0

 

1.5 + -0.5x + -1.5 + x2 = 0 + -1

 

1.5 + -1.5 + -0.5x + x2 = 0 + -1.5

 

 1.5 + -1.5 = 0.0

0.0 + -0.5x + x2 = 0 + -1.5

-0.5x + x2 = 0 + -1.5

 

0 + -1.5 = -1.5

-0.5x + x2 = -1.5

 

-0.5x + 0.0625 + x2 = -1.5 + 0.0625

0.0625 + -0.5x + x2 = -1.5 + 0.0625

 

 -1.5 + 0.0625 = -1.4375

0.0625 + -0.5x + x2 = -1.4375

(x + -0.25)(x + -0.25) = -1.4375

 

x = {0}

Answer 2

a)4x^3 +8x^2-9x-18=0
4x^2(x+2)- 9(x+2)=0
(4x^2-9) (x+ 2) = 0
4x^2 - 9=0 v x+2= 0
x= 3/2 v x=-3/2 v x=-2

 

d) x^4+x^3-7x^2-x+6=0
z schematu Hornera dla1 : (jak na obrazku)
x^3 + 2x^2 -5x-6=0
z schematu Hornera dla 2:
x^2 + 4x+3=0
Δ=4
√Δ=2
x= (-4-2)/2 = -3 v x= -1

 

e)x^3-7x+6=0
z schematu Hornera-->
x^2 -6x+6=0
Δ=12
√Δ= √12
x₁= (6-√12)/2 v x₂= (6+ √12)/2

 

 

g) x^4-3x^3+4x^2-6x+4=0
z schematu Hornera dla 1 -->
x^3 -2x^2 +2x -4=0
z schematu Hornera dla 2 -->
x^2 +2=0
x= √2 v x= -√2

 

Dziękuje za moją pomoc :)