Question

In the given polynomial x^2-2x-√5, find the values of
(I) alpha- beta
(II)alpha/beta+beta/alpha
(III)(alpha^2-beta^2)-(alpha-beta)^2-(alpha+beta)^2
where alpha and beta are the zeroes of the polynomial.

Answer 1

Given polynomial,
x²-2x-√5=
let A and B be the zeroes of given polynomial.
Quadratic formula:
A=[-b+√(b²-4ac)]/2a

B=[-b-√(b²-4ac)]/2a
Now by substituting values we get,
A=[2+√(4+4√5)]/2

A=[2+√4(1+√5)]/2

A=[2+2√(1+√5)]/2

A=[1+√(1+√5)]
and,
B=[1-√(1+√5)]
Now,
sum of the zeroes=A+B=2

A+B=2........................................1
and,
product of zeroesAB=-√5

AB=-√5.......................................2

(i)
=A-B

=1+√(1+√5)-[1-√(1+√5)]

=1+√(1+√5)-1+√(1+√5)

=2√(1+√5)


(ii)
=A/B+B/A

=(A²+B²)/AB

=[(A+B)²-2AB]/AB
By putting eq1 and eq2 we get,
=[(2)²-2(-√5)]/(-√5)

=[4+2√5]/√5


(iii)
=(A²-B²)-(A-B)²-(A+B)²

=A²-B²-(A²+B²-2AB)-(A²+B²+2AB)

=A²-B²-A²-B²+2AB-A²-B²-2AB

=-A²-3B²

=-A²-B²-2B²

=-(A²+B²)-2B²

=-[(A+B)²-2AB]-2B²

=-[4+2√5]-2[1-√(1+√5)]²

=-4-2√5-2[1+1+√5-2√(1+√5)]

=-4-2√5-4-2√5+4√(1+√5)

=4√(1+√5)-8-4√5

Answer 2

1. 2 under root 1 +root5