Question

If a=√3-√2/√3+√2 and b=√3+√2/√3-√2 find a²+b²+ab.

Plz I want answers as fast as possible!!

Answer 1

a = (√3 - √2)/(√3 + √2)
=(√3 -√2)(√3 -√2)/(√3+√2)(√3 - √2)
=(√3 - √2)²/(√3² -√2²)
=(5 -2√6)/1
=( 5 -2√6)

similarly
b = ( √3 + √2)/(√3 - √2)
=( √3 +√2)(√3 +√2)/(√3 -√2)(√3 +√2)
=(√3 +√2)²/(√3² -√2²)
=(5+2√6)/1
=(5 +2√6)

now,

a² + b² + ab
=(a + b)² -2ab + ab
=(a + b)² - ab
={ (5 -2√6)+(5 +2√6)}² - (5-2√6)(5+2√6)
=(10)² - (25 -24)
= 100 - 1
= 99

Answer 2

hi friend,

If a=√3-√2/√3+√2

multiply and divide by √3-√2

a=(√3-√2)²/(3-2)

a=3+2-2√6

a=5-2√6

→b=√3+√2/√3-√2

multiply and divide by √3+√2

→b=(√3+√2)²/(3-2)

→b=5+2√6

now, a²+b²+ab

→(5-2√6)²+(5+2√6)²+1

→25+24+25+24+1

→99


I hope this will help u ;)