Step-by-step explanation:
Given :-
x = (√3-√2)/(√3+√2) and
y = (√3+√2)/(√3-√2)
To find :-
Find the value of x²+y²+xy ?
Solution:-
Given that
x = (√3-√2)/(√3+√2)
The denominator = √3+√2
The Rationalising factor of √3+√2 is √3-√2
On Rationalising the denominator then
=>x=[(√3-√2)/(√3+√2)]×[(√3-√2)/(√3-√2)]
=> x = [(√3-√2)(√3-√2)]×[(√3+√2)(√3-√2)]
=> x = (√3-√2)²/[(√3+√2)(√3-√2)]
=> x = (√3-√2)²/[(√3)²-(√2)²]
Since (a+b)(a-b) = a²-b²
Where , a = √3 and b = √2
=> x = (√3-√2)²/(3-2)
=> x = (√3-√2)²/1
=> x = (√3-√2)²
=> x = (√3)²-2(√3)(√2)+(√2)²
Since , (a-b)² = a²-2ab+b²
Where , a = √3 and b = √2
=> x = 3-2√6+2
=> x = 5-2√6 --------------------(1)
On squaring both sides then
=> x² = (5-2√6)²
=> x² = 5²-2(5)(2√6)+(2√6)²
Since ,(a-b)² = a²-2ab+b²
Where , a = 5 and b = 2√6
=> x² = 25-20√6+24
=> x² = 49-20√6 ---------------(2)
and
y = (√3+√2)/(√3-√2)
The denominator = √3-√2
The Rationalising factor of √3-√2 is √3+√2
On Rationalising the denominator then
=>y=[(√3+√2)/(√3-√2)]×[(√3+√2)/(√3+√2)]
=>y=[(√3+√2)(√3+√2)]×[(√3-√2)(√3+√2)]
=> y = (√3+√2)²/[(√3-√2)(√3+√2)]
=> y = (√3+√2)²/[(√3)²-(√2)²]
Since (a+b)(a-b) = a²-b²
Where , a = √3 and b = √2
=> y = (√3+√2)²/(3-2)
=> y = (√3-√2)²/1
=> y = (√3+√2)²
=> y = (√3)²+2(√3)(√2)+(√2)²
Since , (a+b)² = a²+2ab+b²
Where , a = √3 and b = √2
=> y = 3+2√6+2
=> y = 5+2√6 --------------------(3)
On squaring both sides then
=> y² = (5+2√6)²
=> y² = 5²+2(5)(2√6)+(2√6)²
Since ,(a+b)² = a²+2ab+b²
Where , a = 5 and b = 2√6
=> y² = 25+20√6+24
=> y² = 49+20√6 ---------------(4)
On multiplying (1) and (3) then
xy = (5-2√6)(5+2√6)
=> xy = (5)²-(2√6)²
Since (a+b)(a-b) = a²-b²
Where , a = 5 and b = 2√6
=> xy = 25-24
=> xy = 1 -------------------------(5)
Now,
On adding (2),(4) and (5)
=> x²+y²+xy
=> 49-20√6+49+20√6+1
=> (49+49+1)+(20√6-20√6)
=> 99+0
=> 99
Therefore, x²+y²+xy = 99
Answer:-
The value of x²+y²+xy for the given problem is 99
Used formulae:-
→ (a+b)² = a²+2ab+b²
→ (a-b)² = a²-2ab+b²
→ (a+b)(a-b) = a²-b²
→ The Rationalising factor of √a+√b is √a-√b
→ The Rationalising factor of √a-√b is √a+√b
