Question 4:
Given:
#n=sqrt(6+sqrt11)+sqrt(6-sqrt11)-sqrt22#
Let,
#sqrt(6+sqrt11)=sqrtp+sqrtq#
Then,
#sqrt(6-sqrt11)=sqrtp-sqrtq#
Squaringand adding
#(6+sqrt11)+(6-sqrt11)=p+q+2sqrt(pq)+p+q-2sqrt(pq)#
#12=2(p+q)#
#p+q=12/2=6#
#p+q=6#
Squaring and subtracting
#(6+sqrt11)-(6-sqrt11)=(p+q+2sqrt(pq))-(p+q-2sqrt(pq))#=
#2sqrt11=4sqrt(pq)#
#sqrt(pq)=(2sqrt11)/4=sqrt(11)/2#
Squaring
#pq=11/4=2.75#
#x^2-Sumx+Product=0#
#x^2-6x+2.75=0#
#x^2-5.5x-0.5x+2.75=0#
#x(x-5.5)-0.5(x-5.5)=0#
#(x-5.5)(x-0.5)=0#
#x-5.5=0tox=5.5#
#x-0.5=0tox=0.5#
One of the roots can be p, other will be q.
Thus,
#sqrt(6+sqrt11)=sqrt5.5+sqrt0.5#
It follows that
#sqrt(6-sqrt11)=sqrt5.5-sqrt0.5#
Now,
#sqrt(6+sqrt11)+sqrt(6-sqrt11)-sqrt22=sqrt5.5+sqrt0.5+sqrt5.5-sqrt0.5-sqrt22#
#=2sqrt5.5-sqrt22#
#=qrt4sqrt5.5=sqrt22#
#=sqrt(4xx5.5)-sqrt22#
#=sqrt22-sqrt22#
#=0#
Thus,
#n=0#
