please solve this guys
Answers
Step-by-step explanation:
Given :-
X = (√5-2)/(√5+2)
Y = (√5+2)/(√5-2)
To find :-
Find the following :
I) X²
ii) Y²
iii) XY
iv) X²+Y²+XY
Solution :-
Given that :-
X = (√5-2)/(√5+2)
The denominator = √5+2
The Rationalising factor of√5+2 is √5-2
On Rationalising the denominator then
=> X = [(√5-2)/(√5+2)]×[(√5-2)/(√5-2)]
=>X = [(√5-2)(√5-2)]/[(√5+2)(√5-2)]
=> X =(√5-2)²/[(√5+2)(√5-2)]
=> X =(√5-2)²/[(√5)²-2²]
Since (a+b)(a-b)=a²-b²
Where , a = √5 and b=2
=> X = (√5-2)²/(5-4)
=> X= (√5-2)²/1
=> X= (√5-2)²
=> X = (√5)²-2(√5)(2)+2²
Since (a-b)² = a²-2ab+b²
=> X = 5-4√5+4
=> X = 9-4√5-------------------(1)
On squaring both sides then
=> X² = (9-4√5)²
=> X² = (9)²-2(9)(4√5)+(4√5)²
=> X² = 81-72√5+80
=> X² = 161-72√5 -----------(2)
And
Y = (√5+2)/(√5-2)
The denominator = √5-2
The Rationalising factor of√5-2 is √5+2
On Rationalising the denominator then
=> Y = [(√5+2)/(√5-2)]×[(√5+2)/(√5+2)]
=>Y = [(√5+2)(√5+2)]/[(√5-2)(√5+2)]
=> Y =(√5+2)²/[(√5-2)(√5+2)]
=> Y =(√5+2)²/[(√5)²-2²]
Since (a+b)(a-b)=a²-b²
Where , a = √5 and b=2
=> Y = (√5+2)²/(5-4)
=> Y= (√5+2)²/1
=> Y= (√5+2)²
=> Y = (√5)²+2(√5)(2)+2²
Since (a+b)² = a²+2ab+b²
=> Y = 5+4√5+4
=> Y = 9+4√5-------------------(3)
On squaring both sides then
=> Y² = (9+4√5)²
=> Y² = (9)²+2(9)(4√5)+(4√5)²
=> Y² = 81+72√5+80
=> Y² = 161+72√5 -----------(4)
On multiplying (1) & (3) then
XY = (9-4√5)(9+4√5)
=> XY = (9)²-(4√5)²
Since (a+b)(a-b)=a²-b²
Where , a = 9 and b=4√5
=> XY = 81-80
=> XY = 1 ----------------------(5)
Now
On adding (2), (4) &(5)
=> X²+XY+Y²
=> (161-72√5) +1+(161+72√5)
=> (161+161+1+)+(72√5-72√5)
=> 323
=> X²+XY+Y²= 323
Answer:-
i) The value of X² = 161-72√5
ii) The value of Y² = 161+72√5
iii) The value of XY = 1
iv) The value of X²+Y²+XY = 323
Used formulae:-
- The Rationalising factor of √a+b is √a-b
- The Rationalising factor of √a-b is √a+b
- (a+b)² = a²+2ab+b²
- (a-b)² = a²-2ab+b²
- (a+b)(a-b)=a²-b²
Answer:
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