If a=√3+1/√3-1and b=√3-1/√3+1, find the value of a^2+ab+b^2/a^2-ab+b^2
Answers
Answer:
This problem can be solved in 2 ways.
1st :-
a=3–√+2–√3–√−2–√
=(3–√+2–√)(3–√+2–√)(3–√−2–√)(3–√+2–√)
=(3–√+2–√)23−2
=((3–√)2+(2–√)2+2(3–√)(2–√)
=3+2+26–√
=5+26–√
b=3–√−2–√3–√+2–√
=(3–√−2–√)(3–√−2–√)(3–√+2–√)(3–√−2–√)
=(3–√−2–√)23−2
=((3–√)2+(2–√)2+−2(3–√)(2–√)
=3+2−26–√
=5−26–√
Now
a2+b2
We know that
(x+y)2=x2+y2+2xy
⟹x2+y2=(x+y)2−2xy
If we take x=a and y=b then
a2+b2
=(a+b)2−2ab
=(5+26–√+5−26–√)2−2((5+26–√)(5−26–√))
=(10)2−2(25−24)
Here we use identity (a+b)(a−b)=a2−b2
=100−2
=98
2nd :-
a=3–√+2–√3–√−2–√
b=3–√−2–√3–√+2–√
Now
ab=3–√+2–√3–√−2–√×3–√−2–√3–√+2–√
=1
a+b=3–√+2–√3–√−2–√+3–√−2–√3–√+2–√
=(3–√+2–√)2+(3–√−2–√)2(3–√+2–√)(3–√−2–√)
=5+26–√+5−26–√3−2
=10
Now
(x+y)2=x2+y2+2xy
⟹x2+y2=(x+y)2−2xy
Replacing x=a,y=b
a2+b2=(a+b)2−2ab
=(10)2−2×1
Replacing (x+y)=10 and xy=1
=100−2
=98
Therefore a2+b2=98
It totally depends on you which method to use but I recommend the second one .
Step-by-step explanation: