polynomials (1 Viewer)

[c0okies]

someone help! i have no idea what to do when i look at these questions Dx

1. z^n - 1 =0 has roots 1, z1, z2, ... , zn-1. Show that (1 - z1)(1 - z2) ... (1 - zn-1) =n.

2. The equation x^3+3px^2+3qx+r=0, where p^2 (doesnot =) q, has a double root. Show that (pq-r)^2 =4(p^2-q)(q^2-pr).

 

[Mountain.Dew]

someone help! i have no idea what to do when i look at these questions Dx

1. z^n - 1 =0 has roots 1, z1, z2, ... , zn-1. Show that (1 - z1)(1 - z2) ... (1 - zn-1) =n.

2. The equation x^3+3px^2+3qx+r=0, where p^2 (doesnot =) q, has a double root. Show that (pq-r)^2 =4(p^2-q)(q^2-pr).

okay, here goes, my 2 cents.

1) z^n - 1 =0 has roots 1, z1, z2, ... , zn-1. THEREFORE, the equation z^n - 1 =0 can be rewritten as: (z - z1)(z - z2) ... (z - zn-1) = 0

Also, z^n - 1 = (z-1)(z^n-1 + z^n-2 + ... z^2 + z + 1) = 0,

so (z-1)(z - z1)(z - z2) ... (z - zn-1) = (z-1)(z^n-1 + z^n-2 + ... z^2 + z + 1)

so (z - z1)(z - z2) ... (z - zn-1) = (z^n-1 + z^n-2 + ... z^2 + z + 1)

now sub z = 1 ==> LHS = (1 - z1)(1 - z2) ... (1 - zn-1), RHS = 1 + 1 + 1 +....+ 1, n times.

therefore, (1 - z1)(1 - z2) ... (1 - zn-1) =n

2) F(x) = x^3+3px^2+3qx+r. it has a double root, SO solve F'(x) = 0 for x. substitute that value for x back into F(x), and do some algebraic fiddling around to get (pq-r)^2 =4(p^2-q)(q^2-pr).

 

[c0okies]

Omg Thanks =D