quick help (1 Viewer)

[ronaldinho]

if (2 + i) is a double root of
x^6 - 4x^5 - x^4 + 24x^3 -5x^2 -100x + 125 = 0 find the other roots

 

[Trebla]

Since co-efficients are real, then 2 - i is also a double root. We have found 4 roots so let the other 2 roots be α and β.
Sum of roots: 2(2 + i) + 2(2 - i) + α + β = 4
.: 8 + α + β = 4
.: α + β = - 4 ---- (1)
Product of roots: (2 + i)²(2 - i)²αβ = 125
25αβ = 125
.: αβ = 5 ---- (2)
Using relations (1) and (2), it can be deduced that α and β are roots of the equation: x² + 4x + 5 = 0
.: x² + 4x + 4 + 1 = 0
(x + 2)² + 1 = 0
(x + 2)² = -1
x + 2 = ± i
x = - 2 ± i
.: roots are 2 + i, 2 + i, 2 - i, 2 - i, - 2 + i, - 2 - i

 

[ronaldinho]

thanks for the detailed proof.. got it now..

last one..

Prove that the polynomial P(x) = x^3 - x + 3 cannot have a root in the form p+qrtr, where p and q are rational and rtr is irrational..

 

[ronaldinho]

where did u get alpha plus beta = 4 from??

1st line.,

 

[Shortbreads]

The proof gets a bit dodgy towards the end... but you get the idea (hopefully .)


P(x) = x^3 - x +3

Let p+ q.sqrt(m) be a root of P(x)

Where p and q are non-zero rational numbers
and m > 0 (so that sqrt(m) is real)

​

0 = P(p + q.sqrt(m))
0 = (p + q.sqrt(m))^3 - (p + q.sqrt(m)) + 3
0 = p^3 + 3(p^2)(q.sqrt(m)) + 3(p)(q^2)(m.sqrt(m)) + (q^3)(m.sqrt(m)) - p - q.sqrt(m) + 3

Since 0 is a rational number:

0 = 3.(p^2).q.sqrt(m) + (q^3).m.sqrt(m) - q.sqrt(m)
0 = (q.sqrt(m))(3(p^2) + (q^2)m -1)
0 = 3(p^2) + (q^2)m - 1 Noting q =/ 0 , m =/ 0
p = sqrt((1 - (q^2)m) / 3)

Since p is rational:

1 - (q^2)m must be divisible by three and greater than 0

But m > o

.: 0 < 1 - (q^2)m < 1

But none of the values in this range is divisble by three.

.: p is irrational.

.: p + q.sqrt(m) is not a soloution to P(x)

 

[Trebla]

where did u get alpha plus beta = 4 from??

1st line.,

Simplifying the expression:
2(2 + i) + 2(2 - i) + α + β = 4
.: 2(2 + i + 2 - i) + α + β = 4
2(4) + α + β = 4
8 + α + β = 4
.: α + β = - 4 ---- (1)

 

[jyu]

thanks for the detailed proof.. got it now..

last one..

Prove that the polynomial P(x) = x^3 - x + 3 cannot have a root in the form p+qrtr, where p and q are rational and rtr is irrational..

Proof by contradiction:

Let p+q sqrt(r) be a root of P(x),

i.e. [p+q sqrt(r)]^3 - [p+q sqrt(r)] + 3 = 0, expand and collect rational and irrational parts to obtain

{p(p^2 +3rq^2 - 1) + 3} + q(3p^2 + rq^2 - 1)sqrt(r) = 0.

.: p(p^2 +3rq^2 - 1) + 3 = 0 ........(1)
and 3p^2 + rq^2 - 1 = 0 ........(2)

From equation (1), p^2 + 3rq^2 = 1 - 3/p. Since p^2 > 0, r > 0 and q^2 > 0,
.: 1 - 3/p > 0, .: p > 3.

From equation (2), q^2 = (1-3p^2)/r. Since p > 3 and r > 0, .: q^2 < 0.

Here we have a contradiction because q^2 > 0.

.: p+q sqrt(r) cannot be a root of P(x).

:wave: