Question

if a= √2+1/√2-1 and b= 1/a find the value of a^2+b^2

Answer 1

Answer:

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Answer 2

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 .

I just hope it was write and like me!!