Question

If p = √3−√2 /√3+√2 and q = √3+√2 /√3−√2 , then find p^2+ q^2.

Answer 1

This problem can be solved in 2 ways.

1st :-

=3‾√+2‾√3‾√−2‾√a=3+23−2

=(3‾√+2‾√)(3‾√+2‾√)(3‾√−2‾√)(3‾√+2‾√)=(3+2)(3+2)(3−2)(3+2)

=(3‾√+2‾√)23−2=(3+2)23−2

=((3‾√)2+(2‾√)2+2(3‾√)(2‾√)=((3)2+(2)2+2(3)(2)

=3+2+26‾√=3+2+26

=5+26‾√=5+26

=3‾√−2‾√3‾√+2‾√b=3−23+2

=(3‾√−2‾√)(3‾√−2‾√)(3‾√+2‾√)(3‾√−2‾√)=(3−2)(3−2)(3+2)(3−2)

=(3‾√−2‾√)23−2=(3−2)23−2

=((3‾√)2+(2‾√)2+−2(3‾√)(2‾√)=((3)2+(2)2+−2(3)(2)

=3+2−26‾√=3+2−26

=5−26‾√=5−26

Now

2+2a2+b2

We know that

(+)2=2+2+2(x+y)2=x2+y2+2xy

⟹2+2=(+)2−2⟹x2+y2=(x+y)2−2xy

If we take =x=a and =y=b then

2+2a2+b2

=(+)2−2=(a+b)2−2ab

=(5+26‾√+5−26‾√)2−2((5+26‾√)(5−26‾√))=(5+26+5−26)2−2((5+26)(5−26))

=(10)2−2(25−24)=(10)2−2(25−24)

Here we use identity (+)(−)=2−2(a+b)(a−b)=a2−b2

=100−2=100−2

=98=98

2nd :-

=3‾√+2‾√3‾√−2‾√a=3+23−2

=3‾√−2‾√3‾√+2‾√b=3−23+2

Now

=3‾√+2‾√3‾√−2‾√×3‾√−2‾√3‾√+2‾√ab=3+23−2×3−23+2

=1=1

+=3‾√+2‾√3‾√−2‾√+3‾√−2‾√3‾√+2‾√a+b=3+23−2+3−23+2

=(3‾√+2‾√)2+(3‾√−2‾√)2(3‾√+2‾√)(3‾√−2‾√)=(3+2)2+(3−2)2(3+2)(3−2)

=5+26‾√+5−26‾√3−2=5+26+5−263−2

=10=10

Now

(+)2=2+2+2(x+y)2=x2+y2+2xy

⟹2+2=(+)2−2⟹x2+y2=(x+y)2−2xy

Replacing =,=x=a,y=b

2+2=(+)2−2a2+b2=(a+b)2−2ab

=(10)2−2×1=(10)2−2×1

Replacing (+)=10(x+y)=10 and =1xy=1

=100−2=100−2

=98=98

Therefore 2+2=98a2+b2=98

It totally depends on you which method to use but I recommend the second one .