Question

wyznacz wartość parametru k.

a) W(x)=2x3+(k2+1)x2+x-k; x+1 (czyli x to -1)

b) W(x)=x3-kx2-(k2+3)x-4; x-4 (czyli x to 4)

Answer 1

a)

 W(x)=2x3+(k2+1)x2+x-k; x+1 (czyli x to -1)

W(-1) = 2*(-1)^3 + (k^2 +1) *(-1)^2 + (-1) -k

W(-1) = -2 + k^2 -1 - k

 

 

k^2 - k - 2

 

delta = (-1)^2 - 4*1(-2)

delta = 1 +8

delta = 9

pierwiastek z delty = 3

        1 -3

k1 = ------- = -1

          2

 

         1+3

k2 = -------- = 2

           2

 

b)

W(4) = 4^3 - 4^2k - (k^2+3)*4 -4

W(4) = 64 - 16k -4k^2 - 12 -4

W(4) = -4k^2 - 16k + 48

 

delta = (-16)^2 - 4*(-4)*48

delta = 256 + 768

delta = 1024

pierwiastek z delty = 32

 

          16 - 32

k1 = ------------ = 2

            2*(-4)

 

 

          16+32

k2 = ----------------- = -6

             2,*(-4)

Answer 2

a)

W(-1)=0

czyli:

2*(-1)³ + (k²+1)*(-1)² + (-1) -k =0

-2 +k²+1-1-k =0

k²-k-2=0

Δ=b²-4ac =1+8=9   √Δ =3

k1 = (-b-√Δ)/2a = (1-3)/2-2/2=-1

k2=(-b+√Δ)/2a = (1+3)/2 = 4/2=2

 

k=-1   ∨   k=2

b)

W(4) = 0

czyli:

4³ - k*4² - (k²+3)*4 - 4 =0

 64 - 16k - 4k²-12-4=0

-4k²-16k+48 =0   /:(-4)

k² -4k -12=0

Δ=16 + 48=64       √Δ=8

k1=(4-8)/2=-4/2=-2

k2=(4+8)/2=12/2=6

 

k=-2   ∨  k=6