find the zeroes of polynomial
x²-2x-√5
Answers
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:
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