Question

find the zeroes of polynomial

x²-2x-√5
​

Answer 1

Answer:

hey friend I am going to help ya so take a chill pill

Step-by-step explanation:

2+√[4+4√5]. 2-√[4+4√5].

___________. ___________

2. 2

these are the zeroes , hope it helped｡◕‿◕｡｡◕‿◕｡

Answer 2

Answer:

Brainly.in

What is your question?

Given polynomial,

x²-2x-√5=

let A and B be the zeroes of given polynomial

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